### What Is The Law Of Probability?

- Marvin Harvey
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The law of probability tells us about the probability of specific events occurring. The law of large numbers states that the more trials you have in an experiment, then the closer you get to an accurate probability. The addition rule deals with the case of or in the probability of events occurring.

#### What is the basic law of probability?

The two basic laws of probability are as follows: If and are two events defined on a sample space. Addition rule: When and are not mutually exclusive: P A o r B = P A + P B – P A a n d B. Special case: When and are mutually exclusive, we get P A a n d B = 0.

#### What is the law of probability give an example?

The Law of Probability Updated April 25, 2017 By Andrea Farkas Probability measures the likelihood of an event occurring. Expressed mathematically, probability equals the number of ways a specified event can occur, divided by the total number of all possible event occurrences.

For example, if you have a bag containing three marbles – one blue marble and two green marbles – the probability of grabbing a blue marble sight unseen is 1/3. There is one possible outcome where the blue marble is selected, but three total possible trial outcomes – blue, green, and green. Using the same math the probability of grabbing a green marble is 2/3.

You can discover the unknown probability of an event through experimentation. Using the previous example, say you do not know the probability of drawing a certain colored marble, but you know there are three marbles in the bag. You perform a trial and draw a green marble.

- You perform another trial and draw another green marble.
- At this point you might claim the bag contains only green marbles, but based on two trials, your prediction is not reliable.
- It is possible the bag contains only green marbles or it could be the other two are red and you selected the only green marble sequentially.

If you perform the same trial 100 times you will probably discover you select a green marble around 66% percent of the time. This frequency mirrors the correct probability more accurately than your first experiment. This is the law of large numbers: the greater the number of trials, the more accurately the frequency of an event’s outcome will mirror its actual probability.

Probability can only range from values 0 to 1. A probability of 0 means there are no possible outcomes for that event. In our previous example, the probability of drawing a red marble is zero. A probability of 1 means the event will occur in each and every trial. The probability of drawing either a green marble or a blue marble is 1.

There are no other possible outcomes. In the bag containing one blue marble and two green ones, the probability of drawing a green marble is 2/3. This is an acceptable number because 2/3 is greater than 0, but less than 1-within the range of acceptable probability values.

- Nowing this, you can apply the law of subtraction, which states if you know the probability of an event, you can accurately state the probability of that event not occurring.
- Nowing the probability of drawing a green marble is 2/3, you can subtract that value from 1 and correctly determine the probability of not drawing a green marble: 1/3.

If you want to find the probability of two events occurring in sequential trials, use the law of multiplication. For example, instead of the previous three-marbled bag, say there is a five-marbled bag. There is one blue marble, two green marbles, and two yellow marbles.

- If you want to find the probability of drawing a blue marble and a green marble, in either order (and without returning the first marble to the bag), find the probability of drawing a blue marble and the probability of drawing a green marble.
- The probability of drawing a blue marble from the bag of five marbles is 1/5.

The probability of drawing a green marble from the remaining set is 2/4, or 1/2. Correctly applying the law of multiplication involves multiplying the two probabilities, 1/5 and 1/2, for a probability of 1/10. This expresses the likelihood of the two events occurring together.

Applying what you know about the law of multiplication, you can determine the probability of only one of two events occurring. The law of addition states the probability of one out of two events occurring is equal to the sum of the probabilities of each event occurring individually, minus the probability of both events occurring.

In the five-marbled bag, say you want to know the probability of drawing either a blue marble or a green marble. Add the probability of drawing a blue marble (1/5) to the probability of drawing a green marble (2/5). The sum is 3/5. In the previous example expressing the law of multiplication, we found the probability of drawing both a blue and green marble is 1/10.

## What is the first law of probability?

Lab 2- Probability Introduction What determines the numbers that are selected in the lottery or whether heads or tails results after a coin flip? We can ask similar questions about biological events. What determines whether chromosome 14 will be pulled to the left side of the cell or the right side during anaphase I of meiosis? The answer is chance.

Despite common understandings of chance as something completely unpredictable, in math and science chance is expressed through probability in precise mathematical terms that place predictable limits on chance events. Specifically, probability is defined as the likelihood of a particular chance event occurring among the total number of equally likely chance events.

Mathematically, probability (P) = For a coin toss, we can calculate the probability that heads will result from one toss. If heads is the number of particular chance events of interest, then the numerator is simply 1. The total number of equally likely events is 2 because tails is just as likely as heads.

Thus, the probability is or 50 percent. We can now ask, What is the probability of obtaining heads if we flip the coin a second time? The First Law of Probability states that the results of one chance event have no effect on the results of subsequent chance events. Thus, the probability of obtaining heads the second time you flip it remains at,

Even if you obtained five heads in a row, the odds of heads resulting from a sixth flip remain at, What if we ask a different question, What is the probability of flipping a coin twice and obtaining two heads (or, equivalently, flipping two coins at the same time)? Notice that this is different from the previous question; no coins have been flipped yet.

- In this case, we turn to the Second Law of Probability, which states that the probability of independent chance events occurring together is the product of the probabilities of the separate events.
- Thus, if the probability of one coin coming up heads is, and the independent likelihood of the second coin coming up heads is, then the likelihood that both will come up heads is x =,

The Third Law of Probability considers what happens with chance events that are mutually exclusive, which means that they are not independent. In other words, if one event occurs then another event cannot occur. This frequently applies to combinations.

For example, we could ask, What is the probability that in three coin flips, two heads and one tail will result? There are three possibilities for how this might occur: Possibility 1: head-head-tail Possibility 2: head-tail-head Possibility 3: tail-head-head If sequence 2 comes up, neither 1 nor 3 can happen, although in advance you have no way of knowing which might happen.

All three sequences are equally likely, however. To calculate this, use the Third Law which states that the probability that any of a number of mutually exclusive events will occur is the sum of their independent probabilities. For the example above: P(1) = x x = 1/8 P(2) = x x = 1/8 P(3) = x x = 1/8 P(total) = 3/8 Thus, the probability of obtaining two heads and one tail in three separate coin flips is 3/8.

- These same rules of probability allow us to calculate the odds of parents conceiving particular numbers of girls or boys or of predicting the likelihood that specific chromosomes will segregate together into the same gamete.
- All are chance events.
- Procedure Part I: Simple Probability 1.
- Assume that two sides of a coin represent two alleles, Q and q, for a single gene in the cells that give rise to sperm.

During meiosis these alleles will segregate at anaphase I, and only one allele will end up in the sperm that ultimately fertilizes an egg. If meiosis and fertilization are followed in 30 heterozygous males, Qq, how many times would you predict the Q allele to segregate into the successful sperm? The q allele? Place your answers under Expected in Table 1.1.2.

- What is the probability that the Q and q alleles will be found in these sperm? 3.
- Designate sides of the coin as Q and q and toss the single coin 30 times.
- Record the number of times the Q and q alleles are found in successful sperm.
- Place your answers in Table 1.1 under.
- Observed and calculate and record the difference between Observed and Expected.

(Use negative numbers as appropriate.) 4. Now consider a mating between two heterozygous individuals, Qq. Meiosis must occur in each, resulting in the formation of egg cells and sperm cells. Those gametes then can fuse to form a zygote, or fertilized egg.

- Using the second law of probability, what is the chance that the fertilized egg will be QQ? What is the probability the fertilized egg will be qq? 5.
- Using the third law of probability, what is the chance that the fertilized egg will be Qq? 6.
- Apply the probabilities calculated above to 40 matings, and predict the number of heterozygotes and homozygotes of each type.

Record these under Expected in Table 1.2.7. Designate the sides of two coins as Q and q and toss them together 40 times, recording the allele combination each time in Table 1.2.8. Complete Table 1.2 to compare your predictions (Expected Number) with the actual results (Observed Numbers).9.

Now lets examine a couple that is planning a large family. They plan to have three children, and they would like to know the probability that all three will be boys; all three girls; two boys/one girl; etc. Determine all the possible combinations and probabilities and record them in Table 1.3. Now predict the expected numbers of each class if 32 couples have three children each.10.

Designate the sides of three coins as boy and girl and toss them together 32 times, recording the offspring combination each time. Calculate the difference between observed and expected. Part II: Binomial Expansion Assume that a couple plans to have five children.

- In this case, it is somewhat tedious to outline and then calculate all the possible combinations because the number of independent events has increased beyond three or four.
- Fortunately, if one knows the number of events and the individual probabilities of each alternative event, the probabilities of various combinations can be calculated directly using binomial expansion.

For a hypothetical couple that plans to have one child, boy is one possible outcome (and we call that a ) and girl is the other alternative (and we call that b ). Because boy or girl is equally likely (and equal to ), the total probability of having a boy or girl is a + b = 1.

For a couple interested in two children, we simply calculate (a + b)(a + b) = 1, or (a + b) 2 = 1. Expanding this gives: a 2 + 2ab + b 2 = 1. Probability of two boys: a 2 = () 2 = Probability of one boy, one girl: 2ab = 2()() = 2/4 = Probability of two girls: b 2 = () 2 = Total: 1 Generally, the formula for binomial expansion is (a + b) n, where n equals the number of independent events.

(Your text, P-108, shows how to set up a simple table for determining the coefficients.) For example, if four children were planned, the expansion would appear thus: (a + b) 4 = a 4 + 4a 3 b + 6a 2 b 2 + 4ab 3 + b 4 This can be interpreted as indicating that there are 4 ways to get three boys and one girl, 4a 3 b (and, likewise, to get one boy and 3 girls, 4ab 3 ), 6 ways to get 2 boys and 2 girls (6a 2 b 2 ), etc.

- The terms in this binomial expansion show directly that the probability of 4 boys is: a 4 = (1/2) 4 = 1/16 Similarly, the probability of 3 boys and 1 girl is: 4a 3 b = 4(1/2) 3 (1/2) = 4/16 1.
- Complete the rest of the combinations to be sure you understand this.2.
- Lets return to the couple interested in five children.

Use binomial expansion to calculate the probability of having five boys; four boys and one girl; and all the other combinations. Complete your work on the back of the Probability Worksheet and hand this in before you leave lab. Part III: Chi Square One problem with tests of probability is that they rarely come out perfectly.

For example, you may have expected the Q allele to end up in the successful sperm 15 times out of 30 trials (Table 1.1). (In other words, your hypothesis was that Q segregates randomly.) But how would you interpret an observed experimental result that indicates 13 times, not 15? Was the coin you flipped not a fair coin (or was the meiotic segregation not random) or was the deviation simply within the scope of chance deviation? Chi square analysis allows us to answer these and similar questions.

In the segregation results you recorded in Table 1.1, there were only two outcome classes, Q and q. Lets assume that you observed 13 Q and 17 q segregating into successful sperm. To calculate chi square, use the equation: (observed value expected value) 2 c 2 = expected value This equation sums () the square of the differences between observed and expected and divides by expected.

The expected, or predicted, numbers for this case of 50 percent probability with 30 trials are 30(.50) = 15 Q and 30(.50) = 15 q. Construct a table of these values: Observed Expected (O-E) (O-E) 2 (O-E) 2 /E 13 15 -2 4 0.266 17 15 2 4 0.266 Total 30 30 0 0.532 = c 2 Generally speaking, when the difference between the expected and observed outcomes is small, c 2 is small, and when the difference is large, c 2 is large.

For c 2 to be truly meaningful, however, we must compare it against a table of probabilities, such as the one shown below. This table shows the probabilities that a particular value of c 2 is due only to chance variation (and not, in our example, to an unfair coin or nonrandom segregation).

To interpret our value of c 2, we must know the degrees of freedom (df) for the experiment. In our example, if Q resulted, then q could not, and vice versa. Thus, there was only one degree of freedom, which is defined as one less than the total number of outcome classes. From the table, we can see that 0.532 and df = 1 falls between 0.455 and 1.642, which correspond to probabilities of 0.50 and 0.20, respectively.

In other words, if we were to repeat this same experiment many times (or observe many instances of meiotic segregations), we would expect this much difference between observed and expected values between 20 and 50 percent of the time due to chance deviation alone.

Chi Square Probability Values Reject Reject | |||||||||||||

DF | P=,99 | ,95 | ,90 | ,75 | ,50 | ,25 | ,10 | ,05 | ,01 | ||||

1 | 0.00016 | 0.00393 | 0.01579 | 0.10153 | 0.45494 | 1.32330 | 2.70554 | 3.84146 | 6.63490 | ||||

2 | 0.02010 | 0.10259 | 0.21072 | 0.57536 | 1.38629 | 2.77259 | 4.60517 | 5.99146 | 9.21034 | ||||

3 | 0.11483 | 0.35185 | 0.58437 | 1.21253 | 2.36597 | 4.10834 | 6.25139 | 7.81473 | 11.3448 | ||||

4 | 0.29711 | 0.71072 | 1.06362 | 1.92256 | 3.35669 | 5.38527 | 7.77944 | 9.48773 | 13.2767 | ||||

5 | 0.55430 | 1.14548 | 1.61031 | 2.67460 | 4.35146 | 6.62568 | 9.23636 | 11.0705 | 15.0862 | ||||

6 | 0.87209 | 1.63538 | 2.20413 | 3.45460 | 5.34812 | 7.84080 | 10.6446 | 12.5915 | 16.8118 | ||||

7 | 1.23904 | 2.16735 | 2.83311 | 4.25485 | 6.34581 | 9.03715 | 12.0170 | 14.0671 | 18.4753 | ||||

8 | 1.64650 | 2.73264 | 3.48954 | 5.07064 | 7.34412 | 10.2188 | 13.3615 | 15.5073 | 20.0902 | ||||

9 | 2.08790 | 3.32511 | 4.16816 | 5.89883 | 8.34283 | 11.3887 | 14.6836 | 16.9189 | 21.6659 | ||||

10 | 2.55821 | 3.94030 | 4.86518 | 6.73720 | 9.34182 | 12.5488 | 15.9871 | 18.3070 | 23.2092 | ||||

11 | 3.05348 | 4.57481 | 5.57778 | 7.58414 | 10.3410 | 13.7006 | 17.2750 | 19.6751 | 24.7249 | ||||

12 | 3.57057 | 5.22603 | 6.30380 | 8.43842 | 11.3403 | 14.8454 | 18.5493 | 21.0260 | 26.2169 | ||||

13 | 4.10692 | 5.89186 | 7.04150 | 9.29907 | 12.3397 | 15.9839 | 19.8119 | 22.3620 | 27.6882 | ||||

14 | 4.66043 | 6.57063 | 7.78953 | 10.1653 | 13.3392 | 17.1169 | 21.0641 | 23.6847 | 29.1412 | ||||

15 | 5.22935 | 7.26094 | 8.54676 | 11.0365 | 14.3388 | 18.2450 | 22.3071 | 24.9957 | 30.5779 | ||||

16 | 5.81221 | 7.96165 | 9.31224 | 11.9122 | 15.3385 | 19.3688 | 23.5418 | 26.2962 | 31.9999 | ||||

17 | 6.40776 | 8.67176 | 10.0851 | 12.7919 | 16.3381 | 20.4886 | 24.7690 | 27.5871 | 33.4086 | ||||

18 | 7.01491 | 9.39046 | 10.8649 | 13.6752 | 17.3379 | 21.6048 | 25.9894 | 28.8693 | 34.8053 | ||||

19 | 7.63273 | 10.1170 | 11.6509 | 14.5620 | 18.3376 | 22.7178 | 27.2035 | 30.1435 | 36.1908 | ||||

20 | 8.26040 | 10.8508 | 12.4426 | 15.4517 | 19.3374 | 23.8276 | 28.4119 | 31.4104 | 37.5662 | ||||

21 | 8.89720 | 11.5913 | 13.2396 | 16.3443 | 20.3372 | 24.9347 | 29.6150 | 32.6705 | 38.9321 | ||||

22 | 9.54249 | 12.3380 | 14.0414 | 17.2396 | 21.3370 | 26.0392 | 30.8132 | 33.9244 | 40.2893 | ||||

23 | 10.1957 | 13.0905 | 14.8479 | 18.1373 | 22.3368 | 27.1413 | 32.0069 | 35.1724 | 41.6384 | ||||

24 | 10.8563 | 13.8484 | 15.6586 | 19.0372 | 23.3367 | 28.2411 | 33.1962 | 36.4150 | 42.9798 | ||||

25 | 11.5239 | 14.6114 | 16.4734 | 19.9393 | 24.3365 | 29.3388 | 34.3815 | 37.6524 | 44.3141 | ||||

26 | 12.1981 | 15.3791 | 17.2918 | 20.8434 | 25.3364 | 30.4345 | 35.5631 | 38.8851 | 45.6416 | ||||

27 | 12.8785 | 16.1514 | 18.1139 | 21.7494 | 26.3363 | 31.5284 | 36.7412 | 40.1132 | 46.9629 | ||||

28 | 13.5647 | 16.9278 | 18.9392 | 22.6571 | 27.3362 | 32.6204 | 37.9159 | 41.3371 | 48.2782 | ||||

29 | 14.2564 | 17.7083 | 19.7677 | 23.5665 | 28.3361 | 33.7109 | 39.0874 | 42.5569 | 49.5878 | ||||

30 | 14.9534 | 18.49266 | 20.59923 | 24.47761 | 29.33603 | 34.79974 | 40.25602 | 43.77297 | 50.8921 |

Assignment: 1. Write the hypotheses that you were testing (perhaps unknowingly) in the experiments that produced the data in tables 1.2 and 1.3.2. On a blank sheet of paper, construct appropriate tables to calculate c 2 for the data in Tables 1.2 and 1.3.3.

- Interpret your c 2 values and comment on whether you should support or reject your hypothesis.
- Probability Worksheet Name: Table 1.1 Segregation of Alleles at Meiosis Outcome Classes Observed (O) Expected (E) Difference (O-E) Q (in successful sperm) q (in successful sperm) Totals 30 Table 1.2 Calculating Probabilities of Zygote Genotypes Outcome Combinations Observed Expected Difference Classes (O) (E) (O-E) Q on both coins QQ (homo.

zyg.) Q on one coin, Qq or qQ q on the other (hetero. zyg.) q on both coins qq (homo. zyg.) Totals 4 40 Table 1.3 Probabilities of Different Offspring Combinations Outcome Combinations Probability Observed Expected Difference Classes (O) (E) (O-E) 3 boys BBB x x 32 x 1/8 = 1/8 = 4 2 boys, 1 girl 1 boy, 2 girls 3 girls Totals Materials: Calculators PC’s with interconnection pile of pennies References: Chi Square Table Values: http://www.statsoft.com/textbook/sttable.html#chi Chi Square Calculator: http://faculty.vassar.edu/lowry/csfit.html

## What are two basic laws of probability?

Probability Topics OpenStaxCollege When calculating probability, there are two rules to consider when determining if two events are independent or dependent and if they are mutually exclusive or not. If A and B are two events defined on a sample space, then: P ( A AND B ) = P ( B ) P ( A | B ).

- This rule may also be written as: P ( A | B ) = \(\frac B\right)} \) (The probability of A given B equals the probability of A and B divided by the probability of B,) If A and B are independent, then P ( A | B ) = P ( A ).
- Then P ( A AND B ) = P ( A | B ) P ( B ) becomes P ( A AND B ) = P ( A ) P ( B ).

If A and B are defined on a sample space, then: P ( A OR B ) = P ( A ) + P ( B ) – P ( A AND B ). If A and B are mutually exclusive, then P ( A AND B ) = 0. Then P ( A OR B ) = P ( A ) + P ( B ) – P ( A AND B ) becomes P ( A OR B ) = P ( A ) + P ( B ).

- Klaus can only afford one vacation. The probability that he chooses A is P ( A ) = 0.6 and the probability that he chooses B is P ( B ) = 0.35.
- P ( A AND B ) = 0 because Klaus can only afford to take one vacation
- Therefore, the probability that he chooses either New Zealand or Alaska is P ( A OR B ) = P ( A ) + P ( B ) = 0.6 + 0.35 = 0.95. Note that the probability that he does not choose to go anywhere on vacation must be 0.05.

Carlos plays college soccer. He makes a goal 65% of the time he shoots. Carlos is going to attempt two goals in a row in the next game. A = the event Carlos is successful on his first attempt. P ( A ) = 0.65. B = the event Carlos is successful on his second attempt.

P ( B ) = 0.65. Carlos tends to shoot in streaks. The probability that he makes the second goal GIVEN that he made the first goal is 0.90.a. What is the probability that he makes both goals? a. The problem is asking you to find P ( A AND B ) = P ( B AND A ). Since P ( B | A ) = 0.90: P ( B AND A ) = P ( B | A ) P ( A ) = (0.90)(0.65) = 0.585 Carlos makes the first and second goals with probability 0.585.b.

What is the probability that Carlos makes either the first goal or the second goal? b. The problem is asking you to find P ( A OR B ). P ( A OR B ) = P ( A ) + P ( B ) – P ( A AND B ) = 0.65 + 0.65 – 0.585 = 0.715 Carlos makes either the first goal or the second goal with probability 0.715.c.

- Are A and B independent? c.
- No, they are not, because P ( B AND A ) = 0.585.
- P ( B ) P ( A ) = (0.65)(0.65) = 0.423 0.423 ≠ 0.585 = P ( B AND A ) So, P ( B AND A ) is not equal to P ( B ) P ( A ).d.
- Are A and B mutually exclusive? d.
- No, they are not because P ( A and B ) = 0.585.
- To be mutually exclusive, P ( A AND B ) must equal zero.

Try It Helen plays basketball. For free throws, she makes the shot 75% of the time. Helen must now attempt two free throws. C = the event that Helen makes the first shot. P ( C ) = 0.75. D = the event Helen makes the second shot. P ( D ) = 0.75. The probability that Helen makes the second free throw given that she made the first is 0.85.

What is the probability that Helen makes both free throws? P ( D | C ) = 0.85 P ( C AND D ) = P ( D AND C ) P ( D AND C ) = P ( D | C ) P ( C ) = (0.85)(0.75) = 0.6375 Helen makes the first and second free throws with probability 0.6375. A community swim team has 150 members. Seventy-five of the members are advanced swimmers.

Forty-seven of the members are intermediate swimmers. The remainder are novice swimmers. Forty of the advanced swimmers practice four times a week. Thirty of the intermediate swimmers practice four times a week. Ten of the novice swimmers practice four times a week.

Suppose one member of the swim team is chosen randomly.a. What is the probability that the member is a novice swimmer? b. What is the probability that the member practices four times a week? c. What is the probability that the member is an advanced swimmer and practices four times a week? d. What is the probability that a member is an advanced swimmer and an intermediate swimmer? Are being an advanced swimmer and an intermediate swimmer mutually exclusive? Why or why not? d.

P (advanced AND intermediate) = 0, so these are mutually exclusive events. A swimmer cannot be an advanced swimmer and an intermediate swimmer at the same time.e. Are being a novice swimmer and practicing four times a week independent events? Why or why not? e.

- No, these are not independent events.
- P (novice AND practices four times per week) = 0.0667 P (novice) P (practices four times per week) = 0.0996 0.0667 ≠ 0.0996 Try It A school has 200 seniors of whom 140 will be going to college next year.
- Forty will be going directly to work.
- The remainder are taking a gap year.

Fifty of the seniors going to college play sports. Thirty of the seniors going directly to work play sports. Five of the seniors taking a gap year play sports. What is the probability that a senior is taking a gap year? \(P=\frac =\frac =0.1\) Felicity attends Modesto JC in Modesto, CA.

- What is the probability that Felicity enrolls in math and speech? Find P ( M AND S ) = P ( M | S ) P ( S ).
- What is the probability that Felicity enrolls in math or speech classes? Find P ( M OR S ) = P ( M ) + P ( S ) – P ( M AND S ).
- Are M and S independent? Is P ( M | S ) = P ( M )?
- Are M and S mutually exclusive? Is P ( M AND S ) = 0?

a.0.1625, b.0.6875, c. No, d. No Try It A student goes to the library. Let events B = the student checks out a book and D = the student check out a DVD. Suppose that P ( B ) = 0.40, P ( D ) = 0.30 and P ( D | B ) = 0.5.

- Find P ( B AND D ).
- Find P ( B OR D ).

- P ( B AND D ) = P ( D | B ) P ( B ) = (0.5)(0.4) = 0.20.
- P ( B OR D ) = P ( B ) + P ( D ) − P ( B AND D ) = 0.40 + 0.30 − 0.20 = 0.50

Studies show that about one woman in seven (approximately 14.3%) who live to be 90 will develop breast cancer. Suppose that of those women who develop breast cancer, a test is negative 2% of the time. Also suppose that in the general population of women, the test for breast cancer is negative about 85% of the time.

Let B = woman develops breast cancer and let N = tests negative. Suppose one woman is selected at random.a. What is the probability that the woman develops breast cancer? What is the probability that woman tests negative? a. P ( B ) = 0.143; P ( N ) = 0.85 b. Given that the woman has breast cancer, what is the probability that she tests negative? c.

What is the probability that the woman has breast cancer AND tests negative? c. P ( B AND N ) = P ( B ) P ( N | B ) = (0.143)(0.02) = 0.0029 d. What is the probability that the woman has breast cancer or tests negative? d. P ( B OR N ) = P ( B ) + P ( N ) – P ( B AND N ) = 0.143 + 0.85 – 0.0029 = 0.9901 e.

Are having breast cancer and testing negative independent events? e. No. P ( N ) = 0.85; P ( N | B ) = 0.02. So, P ( N | B ) does not equal P ( N ).f. Are having breast cancer and testing negative mutually exclusive? f. No. P ( B AND N ) = 0.0029. For B and N to be mutually exclusive, P ( B AND N ) must be zero.

Try It A school has 200 seniors of whom 140 will be going to college next year. Forty will be going directly to work. The remainder are taking a gap year. Fifty of the seniors going to college play sports. Thirty of the seniors going directly to work play sports.

- Five of the seniors taking a gap year play sports.
- What is the probability that a senior is going to college and plays sports? Let A = student is a senior going to college.
- Let B = student plays sports.
- P ( B ) = \(\frac \) P ( B | A ) = \(\frac \) P ( A AND B ) = P ( B | A ) P ( A ) P ( A AND B ) = \(\left(\frac \right)\left(\frac \right)\) = \(\frac \) Refer to the information in,

P = tests positive.

- Given that a woman develops breast cancer, what is the probability that she tests positive. Find P ( P | B ) = 1 – P ( N | B ).
- What is the probability that a woman develops breast cancer and tests positive. Find P ( B AND P ) = P ( P | B ) P ( B ).
- What is the probability that a woman does not develop breast cancer. Find P ( B′ ) = 1 – P ( B ).
- What is the probability that a woman tests positive for breast cancer. Find P ( P ) = 1 – P ( N ).

a.0.98; b.0.1401; c.0.857; d.0.15 Try It A student goes to the library. Let events B = the student checks out a book and D = the student checks out a DVD. Suppose that P ( B ) = 0.40, P ( D ) = 0.30 and P ( D | B ) = 0.5.

- Find P ( B′ ).
- Find P ( D AND B ).
- Find P ( B | D ).
- Find P ( D AND B′ ).
- Find P ( D | B′ ).

- P ( B′ ) = 0.60
- P ( D AND B ) = P ( D | B ) P ( B ) = 0.20
- P ( B | D ) = \(\frac B\text D\text } D\text }\) = \(\frac \text }\) = 0.66
- P ( D AND B′ ) = P ( D ) – P ( D AND B ) = 0.30 – 0.20 = 0.10
- P ( D | B′ ) = P ( D AND B′ ) P ( B′ ) = ( P ( D ) – P ( D AND B ))(0.60) = (0.10)(0.60) = 0.06

DiCamillo, Mark, Mervin Field. “The File Poll.” Field Research Corporation. Available online at http://www.field.com/fieldpollonline/subscribers/Rls2443.pdf (accessed May 2, 2013). Rider, David, “Ford support plummeting, poll suggests,” The Star, September 14, 2011.

- Available online at http://www.thestar.com/news/gta/2011/09/14/ford_support_plummeting_poll_suggests.html (accessed May 2, 2013).
- Mayor’s Approval Down.” News Release by Forum Research Inc.
- Available online at http://www.forumresearch.com/forms/News Archives/News Releases/74209_TO_Issues_-_Mayoral_Approval_%28Forum_Research%29%2820130320%29.pdf (accessed May 2, 2013).

“Roulette.” Wikipedia. Available online at http://en.wikipedia.org/wiki/Roulette (accessed May 2, 2013). Shin, Hyon B., Robert A. Kominski. “Language Use in the United States: 2007.” United States Census Bureau. Available online at http://www.census.gov/hhes/socdemo/language/data/acs/ACS-12.pdf (accessed May 2, 2013).

Data from the Baseball-Almanac, 2013. Available online at www.baseball-almanac.com (accessed May 2, 2013). Data from U.S. Census Bureau. Data from the Wall Street Journal. Data from The Roper Center: Public Opinion Archives at the University of Connecticut. Available online at http://www.ropercenter.uconn.edu/ (accessed May 2, 2013).

### Laws of Probability (Basics)

Data from Field Research Corporation. Available online at www.field.com/fieldpollonline (accessed May 2,2 013). The multiplication rule and the addition rule are used for computing the probability of A and B, as well as the probability of A or B for two given events A, B defined on the sample space.

In sampling with replacement each member of a population is replaced after it is picked, so that member has the possibility of being chosen more than once, and the events are considered to be independent. In sampling without replacement, each member of a population may be chosen only once, and the events are considered to be not independent.

The events A and B are mutually exclusive events when they do not have any outcomes in common. The multiplication rule: P ( A AND B ) = P ( A | B ) P ( B ) The addition rule: P ( A OR B ) = P ( A ) + P ( B ) – P ( A AND B ) Use the following information to answer the next ten exercises.

- Forty-eight percent of all Californians registered voters prefer life in prison without parole over the death penalty for a person convicted of first degree murder.
- Among Latino California registered voters, 55% prefer life in prison without parole over the death penalty for a person convicted of first degree murder.37.6% of all Californians are Latino.

In this problem, let: C = Californians (registered voters) preferring life in prison without parole over the death penalty for a person convicted of first degree murder. L = Latino Californians Suppose that one Californian is randomly selected. In words, what is C | L ? C | L means, given the person chosen is a Latino Californian, the person is a registered voter who prefers life in prison without parole for a person convicted of first degree murder.

In words, what is L AND C ? L AND C is the event that the person chosen is a Latino California registered voter who prefers life without parole over the death penalty for a person convicted of first degree murder. Are L and C independent events? Show why or why not. In words, what is L OR C ? Are L and C mutually exclusive events? Show why or why not.

No, because P ( L AND C ) does not equal 0. On February 28, 2013, a Field Poll Survey reported that 61% of California registered voters approved of allowing two people of the same gender to marry and have regular marriage laws apply to them. Among 18 to 39 year olds (California registered voters), the approval rating was 78%.

- C = California registered voters who support same-sex marriage.
- B = California registered voters who say the Supreme Court’s ruling about the constitutionality of California’s Proposition 8 is very or somewhat important to them
- A = California registered voters who are 18 to 39 years old.

- Find P ( C ).
- Find P ( B ).
- Find P ( C | A ).
- Find P ( B | C ).
- In words, what is C | A ?
- In words, what is B | C ?
- Find P ( C AND B ).
- In words, what is C AND B ?
- Find P ( C OR B ).
- Are C and B mutually exclusive events? Show why or why not.

After Rob Ford, the mayor of Toronto, announced his plans to cut budget costs in late 2011, the Forum Research polled 1,046 people to measure the mayor’s popularity. Everyone polled expressed either approval or disapproval. These are the results their poll produced:

- In early 2011, 60 percent of the population approved of Mayor Ford’s actions in office.
- In mid-2011, 57 percent of the population approved of his actions.
- In late 2011, the percentage of popular approval was measured at 42 percent.

- What is the sample size for this study?
- What proportion in the poll disapproved of Mayor Ford, according to the results from late 2011?
- How many people polled responded that they approved of Mayor Ford in late 2011?
- What is the probability that a person supported Mayor Ford, based on the data collected in mid-2011?
- What is the probability that a person supported Mayor Ford, based on the data collected in early 2011?

- The Forum Research surveyed 1,046 Torontonians.
- 58%
- 42% of 1,046 = 439 (rounding to the nearest integer)
- 0.57
- 0.60.

Use the following information to answer the next three exercises. The casino game, roulette, allows the gambler to bet on the probability of a ball, which spins in the roulette wheel, landing on a particular color, number, or range of numbers. The table used to place bets contains of 38 numbers, and each number is assigned to a color and a range. (credit: film8ker/wikibooks)

- List the sample space of the 38 possible outcomes in roulette.
- You bet on red. Find P (red).
- You bet on -1st 12- (1st Dozen). Find P (-1st 12-).
- You bet on an even number. Find P (even number).
- Is getting an odd number the complement of getting an even number? Why?
- Find two mutually exclusive events.
- Are the events Even and 1st Dozen independent?

Compute the probability of winning the following types of bets:

- Betting on two lines that touch each other on the table as in 1-2-3-4-5-6
- Betting on three numbers in a line, as in 1-2-3
- Betting on one number
- Betting on four numbers that touch each other to form a square, as in 10-11-13-14
- Betting on two numbers that touch each other on the table, as in 10-11 or 10-13
- Betting on 0-00-1-2-3
- Betting on 0-1-2; or 0-00-2; or 00-2-3

- P (Betting on two line that touch each other on the table) = \(\frac \)
- P (Betting on three numbers in a line) = \(\frac \)
- P (Bettting on one number) = \(\frac \)
- P (Betting on four number that touch each other to form a square) = \(\frac \)
- P (Betting on two number that touch each other on the table ) = \(\frac \)
- P (Betting on 0-00-1-2-3) = \(\frac \)
- P (Betting on 0-1-2; or 0-00-2; or 00-2-3) = \(\frac \)

Compute the probability of winning the following types of bets:

- Betting on a color
- Betting on one of the dozen groups
- Betting on the range of numbers from 1 to 18
- Betting on the range of numbers 19–36
- Betting on one of the columns
- Betting on an even or odd number (excluding zero)

Suppose that you have eight cards. Five are green and three are yellow. The five green cards are numbered 1, 2, 3, 4, and 5. The three yellow cards are numbered 1, 2, and 3. The cards are well shuffled. You randomly draw one card.

- G = card drawn is green
- E = card drawn is even-numbered
- List the sample space.
- P ( G ) = _
- P ( G | E ) = _
- P ( G AND E ) = _
- P ( G OR E ) = _
- Are G and E mutually exclusive? Justify your answer numerically.

- \(\frac \text \)
- \(\frac \text \)
- \(\frac \text \)
- \(\frac \text \)
- No, because P ( G AND E ) does not equal 0.

Roll two fair dice. Each die has six faces.

- List the sample space.
- Let A be the event that either a three or four is rolled first, followed by an even number. Find P ( A ).
- Let B be the event that the sum of the two rolls is at most seven. Find P ( B ).
- In words, explain what ” P ( A | B )” represents. Find P ( A | B ).
- Are A and B mutually exclusive events? Explain your answer in one to three complete sentences, including numerical justification.
- Are A and B independent events? Explain your answer in one to three complete sentences, including numerical justification.

A special deck of cards has ten cards. Four are green, three are blue, and three are red. When a card is picked, its color of it is recorded. An experiment consists of first picking a card and then tossing a coin.

- List the sample space.
- Let A be the event that a blue card is picked first, followed by landing a head on the coin toss. Find P ( A ).
- Let B be the event that a red or green is picked, followed by landing a head on the coin toss. Are the events A and B mutually exclusive? Explain your answer in one to three complete sentences, including numerical justification.
- Let C be the event that a red or blue is picked, followed by landing a head on the coin toss. Are the events A and C mutually exclusive? Explain your answer in one to three complete sentences, including numerical justification.

NOTE The coin toss is independent of the card picked first.

- P ( A ) = P (blue) P (head) = \(\left(\frac \right)\)\(\left(\frac \right)\) = \(\frac \)
- Yes, A and B are mutually exclusive because they cannot happen at the same time; you cannot pick a card that is both blue and also (red or green). P ( A AND B ) = 0
- No, A and C are not mutually exclusive because they can occur at the same time. In fact, C includes all of the outcomes of A ; if the card chosen is blue it is also (red or blue). P ( A AND C ) = P ( A ) = \(\frac \)

An experiment consists of first rolling a die and then tossing a coin.

- List the sample space.
- Let A be the event that either a three or a four is rolled first, followed by landing a head on the coin toss. Find P ( A ).
- Let B be the event that the first and second tosses land on heads. Are the events A and B mutually exclusive? Explain your answer in one to three complete sentences, including numerical justification.

An experiment consists of tossing a nickel, a dime, and a quarter. Of interest is the side the coin lands on.

- List the sample space.
- Let A be the event that there are at least two tails. Find P ( A ).
- Let B be the event that the first and second tosses land on heads. Are the events A and B mutually exclusive? Explain your answer in one to three complete sentences, including justification.

- S =
- \(\frac \)
- Yes, because if A has occurred, it is impossible to obtain two tails. In other words, P ( A AND B ) = 0.

Consider the following scenario: Let P ( C ) = 0.4. Let P ( D ) = 0.5. Let P ( C | D ) = 0.6. Find P ( C AND D ). Are C and D mutually exclusive? Why or why not? Are C and D independent events? Why or why not? Find P ( C OR D ). Find P ( D | C ). Y and Z are independent events.

- Rewrite the basic Addition Rule P ( Y OR Z ) = P ( Y ) + P ( Z ) – P ( Y AND Z ) using the information that Y and Z are independent events.
- Use the rewritten rule to find P ( Z ) if P ( Y OR Z ) = 0.71 and P ( Y ) = 0.42.

- If Y and Z are independent, then P ( Y AND Z ) = P ( Y ) P ( Z ), so P ( Y OR Z ) = P ( Y ) + P ( Z ) – P ( Y ) P ( Z ).
- 0.5

G and H are mutually exclusive events. P ( G ) = 0.5 P ( H ) = 0.3

- Explain why the following statement MUST be false: P ( H | G ) = 0.4.
- Find P ( H OR G ).
- Are G and H independent or dependent events? Explain in a complete sentence.

Approximately 281,000,000 people over age five live in the United States. Of these people, 55,000,000 speak a language other than English at home. Of those who speak another language at home, 62.3% speak Spanish. Let: E = speaks English at home; E′ = speaks another language at home; S = speaks Spanish; Finish each probability statement by matching the correct answer.

Probability Statements | Answers |
---|---|

a. P ( E′ ) = | i.0.8043 |

b. P ( E ) = | ii.0.623 |

c. P ( S and E′ ) = | iii.0.1957 |

d. P ( S | E′ ) = | iv.0.1219 |

iii i iv ii 1994, the U.S. government held a lottery to issue 55,000 Green Cards (permits for non-citizens to work legally in the U.S.). Renate Deutsch, from Germany, was one of approximately 6.5 million people who entered this lottery. Let G = won green card.

- What was Renate’s chance of winning a Green Card? Write your answer as a probability statement.
- In the summer of 1994, Renate received a letter stating she was one of 110,000 finalists chosen. Once the finalists were chosen, assuming that each finalist had an equal chance to win, what was Renate’s chance of winning a Green Card? Write your answer as a conditional probability statement. Let F = was a finalist.
- Are G and F independent or dependent events? Justify your answer numerically and also explain why.
- Are G and F mutually exclusive events? Justify your answer numerically and explain why.

Three professors at George Washington University did an experiment to determine if economists are more selfish than other people. They dropped 64 stamped, addressed envelopes with 💲10 cash in different classrooms on the George Washington campus.44% were returned overall.

- Write a probability statement for the overall percent of money returned.
- Write a probability statement for the percent of money returned out of the economics classes.
- Write a probability statement for the percent of money returned out of the other classes.
- Is money being returned independent of the class? Justify your answer numerically and explain it.
- Based upon this study, do you think that economists are more selfish than other people? Explain why or why not. Include numbers to justify your answer.

- P ( R ) = 0.44
- P ( R | E ) = 0.56
- P ( R | O ) = 0.31
- No, whether the money is returned is not independent of which class the money was placed in. There are several ways to justify this mathematically, but one is that the money placed in economics classes is not returned at the same overall rate; P ( R | E ) ≠ P ( R ).
- No, this study definitely does not support that notion; in fact, it suggests the opposite. The money placed in the economics classrooms was returned at a higher rate than the money place in all classes collectively; P ( R | E ) > P ( R ).

The following table of data obtained from www.baseball-almanac.com shows hit information for four players. Suppose that one hit from the table is randomly selected.

Name | Single | Double | Triple | Home Run | Total Hits |
---|---|---|---|---|---|

Babe Ruth | 1,517 | 506 | 136 | 714 | 2,873 |

Jackie Robinson | 1,054 | 273 | 54 | 137 | 1,518 |

Ty Cobb | 3,603 | 174 | 295 | 114 | 4,189 |

Hank Aaron | 2,294 | 624 | 98 | 755 | 3,771 |

Total | 8,471 | 1,577 | 583 | 1,720 | 12,351 |

Are “the hit being made by Hank Aaron” and “the hit being a double” independent events?

- Yes, because P (hit by Hank Aaron|hit is a double) = P (hit by Hank Aaron)
- No, because P (hit by Hank Aaron|hit is a double) ≠ P (hit is a double)
- No, because P (hit is by Hank Aaron|hit is a double) ≠ P (hit by Hank Aaron)
- Yes, because P (hit is by Hank Aaron|hit is a double) = P (hit is a double)

United Blood Services is a blood bank that serves more than 500 hospitals in 18 states. According to their website, a person with type O blood and a negative Rh factor (Rh-) can donate blood to any person with any bloodtype. Their data show that 43% of people have type O blood and 15% of people have Rh- factor; 52% of people have type O or Rh- factor.

- Find the probability that a person has both type O blood and the Rh- factor.
- Find the probability that a person does NOT have both type O blood and the Rh- factor.

- P (type O OR Rh-) = P (type O) + P (Rh-) – P (type O AND Rh-) 0.52 = 0.43 + 0.15 – P (type O AND Rh-); solve to find P (type O AND Rh-) = 0.06 6% of people have type O, Rh- blood
- P (NOT(type O AND Rh-)) = 1 – P (type O AND Rh-) = 1 – 0.06 = 0.94 94% of people do not have type O, Rh- blood

At a college, 72% of courses have final exams and 46% of courses require research papers. Suppose that 32% of courses have a research paper and a final exam. Let F be the event that a course has a final exam. Let R be the event that a course requires a research paper.

- Find the probability that a course has a final exam or a research project.
- Find the probability that a course has NEITHER of these two requirements.

In a box of assorted cookies, 36% contain chocolate and 12% contain nuts. Of those, 8% contain both chocolate and nuts. Sean is allergic to both chocolate and nuts.

- Find the probability that a cookie contains chocolate or nuts (he can’t eat it).
- Find the probability that a cookie does not contain chocolate or nuts (he can eat it).

- Let C = be the event that the cookie contains chocolate. Let N = the event that the cookie contains nuts.
- P ( C OR N ) = P ( C ) + P ( N ) – P ( C AND N ) = 0.36 + 0.12 – 0.08 = 0.40
- P (NEITHER chocolate NOR nuts) = 1 – P ( C OR N ) = 1 – 0.40 = 0.60

A college finds that 10% of students have taken a distance learning class and that 40% of students are part time students. Of the part time students, 20% have taken a distance learning class. Let D = event that a student takes a distance learning class and E = event that a student is a part time student

- Find P ( D AND E ).
- Find P ( E | D ).
- Find P ( D OR E ).
- Using an appropriate test, show whether D and E are independent.
- Using an appropriate test, show whether D and E are mutually exclusive.

#### What is probability and its formula?

What is the probability formula? The formula for probability states that the possibility of an event happening, or P(E), equals the ratio of the number of favorable outcomes to the number of total outcomes. Mathematically, it looks like this: P(E) = favorable outcomes/total outcomes.

#### What are 5 rules of probability?

Lesson Plan – Probability is a measure of the likelihood of an event to occur. Many events cannot be predicted with total certainty. Using probability, one can predict only the chance of an event to occur, i.e., how likely they are going to happen. For example, when a coin is tossed, there is a probability to get heads or tails. Properties:

- Probability of an impossible event is phi or a null set.
- The maximum probability of an event is its sample space (sample space is the total number of possible outcomes)
- Probability of any event exists between 0 and 1. (0 can also be a probability).
- There cannot be a negative probability for an event.
- If A and B are two mutually exclusive outcomes (Two events that cannot occur at the same time), then the probability of A or B occurring is the probability of A plus the probability of B.

The probability formula is the ratio of the possibility of occurrence of an outcome to the total number of outcomes. Probability of occurrence of an event P(E) = Number of favorable outcomes/Total Number of outcomes.

### Who gave law of probability?

A Short History of Probability From Calculus, Volume II by (2 nd edition, John Wiley & Sons, 1969 ): “A gambler’s dispute in 1654 led to the creation of a mathematical theory of probability by two famous French mathematicians, Blaise Pascal and Pierre de Fermat.

- Antoine Gombaud, Chevalier de Méré, a French nobleman with an interest in gaming and gambling questions, called Pascal’s attention to an apparent contradiction concerning a popular dice game.
- The game consisted in throwing a pair of dice 24 times; the problem was to decide whether or not to bet even money on the occurrence of at least one “double six” during the 24 throws.

A seemingly well-established gambling rule led de Méré to believe that betting on a double six in 24 throws would be profitable, but his own calculations indicated just the opposite. This problem and others posed by de Méré led to an exchange of letters between Pascal and Fermat in which the fundamental principles of probability theory were formulated for the first time.

- Although a few special problems on games of chance had been solved by some Italian mathematicians in the 15th and 16th centuries, no general theory was developed before this famous correspondence.
- The Dutch scientist Christian Huygens, a teacher of Leibniz, learned of this correspondence and shortly thereafter (in 1657) published the first book on probability; entitled De Ratiociniis in Ludo Aleae, it was a treatise on problems associated with gambling.

Because of the inherent appeal of games of chance, probability theory soon became popular, and the subject developed rapidly during the 18th century. The major contributors during this period were Jakob Bernoulli (1654-1705) and Abraham de Moivre (1667-1754).

In 1812 Pierre de Laplace (1749-1827) introduced a host of new ideas and mathematical techniques in his book, Théorie Analytique des Probabilités, Before Laplace, probability theory was solely concerned with developing a mathematical analysis of games of chance. Laplace applied probabilistic ideas to many scientific and practical problems.

The theory of errors, actuarial mathematics, and statistical mechanics are examples of some of the important applications of probability theory developed in the l9th century. Like so many other branches of mathematics, the development of probability theory has been stimulated by the variety of its applications.

Conversely, each advance in the theory has enlarged the scope of its influence. Mathematical statistics is one important branch of applied probability; other applications occur in such widely different fields as genetics, psychology, economics, and engineering. Many workers have contributed to the theory since Laplace’s time; among the most important are Chebyshev, Markov, von Mises, and Kolmogorov.

One of the difficulties in developing a mathematical theory of probability has been to arrive at a definition of probability that is precise enough for use in mathematics, yet comprehensive enough to be applicable to a wide range of phenomena. The search for a widely acceptable definition took nearly three centuries and was marked by much controversy.

#### Who discovered law of probability?

History of probability – The modern mathematical theory of probability has its roots in attempts to analyze games of chance by Gerolamo Cardano in the sixteenth century, and by Pierre de Fermat and Blaise Pascal in the seventeenth century (for example the ” problem of points “).

Christiaan Huygens published a book on the subject in 1657. In the 19th century, what is considered the classical definition of probability was completed by Pierre Laplace, Initially, probability theory mainly considered discrete events, and its methods were mainly combinatorial, Eventually, analytical considerations compelled the incorporation of continuous variables into the theory.

This culminated in modern probability theory, on foundations laid by Andrey Nikolaevich Kolmogorov, Kolmogorov combined the notion of sample space, introduced by Richard von Mises, and measure theory and presented his axiom system for probability theory in 1933.

## What are the first three laws of probability?

There are three main rules associated with basic probability: the addition rule, the multiplication rule, and the complement rule.

### How many theory are there in probability?

What are the Two Types of Probabilities in Probability Theory? – The two types of probabilities in probability theory are theoretical probability and experimental probability. Theoretical probability gives the probability of what is expected to happen without conducting any experiments. Experimental probability uses repeated experiments to give the probability of an event taking place.

### What are the terms of probability?

6. What are the 3 axioms of probability? – The three axioms of probability are:

- The probability of an event is always a real number between 0 and 1.
- The sum of the probabilities of all the events in an experiment is 1.
- The happening of either of the two independent events is equal to the sum of their individual probabilities. \

## What is the law of probability distribution?

What Is a Probability Distribution? – A probability distribution is a statistical function that describes all the possible values and likelihoods that a random variable can take within a given range. This range will be bounded between the minimum and maximum possible values, but precisely where the possible value is likely to be plotted on the probability distribution depends on a number of factors.

### What are the first three laws of probability?

There are three main rules associated with basic probability: the addition rule, the multiplication rule, and the complement rule.