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Probability Exercises.
Ma 162 Spring 2010
Ma 162 Spring 2010
April 21, 2010
Problem 1.
(^) Conditional Probability: It is known that a student who does his online homework on a regular basis has a chance of 83 percent to get a good grade (A or B); but the chance drops to 58 percent if he doesn’t do the homework regularly.
(^) John has been very busy with other courses and an evening job and figures that he has only a 69 percent chance of doing the homework regularly. What is his chance of not getting a good grade in the course?
Answer: There are two important events under discussion. Let us name them.
(^) E: doing regular homework regularly. F: getting a good grade.
Problem 1 Continued.
There are two events E (regular homework) and F (good grade).
(^) We are given that P (F |E) = 0.83 and P (F |EC^ ) = 0.58. We are also given that P (E) = 0.69 and hence P (EC^ ) = 1 − 0 .69 = 0.31.
(^) We get four possible paths to final outcome.
(^) E happens and then F happens. Thus E ⋂ F happens. (^) E happens and F does not happen, i.e. E ⋂ F C^ happens. E does not happen and F happens. Thus EC^
⋂ F happens. Finally, neither E nor F happen, i.e. EC^
⋂ F C^ happens.
Problem 1 calculations.
(^) Note that: P (E
F ) = P (E) · P (F |E) = (0.69) · (0.83) = 0.5727.
(^) We note that P (F C^ |E) = 1 − P (F |E) = 0.17. So P (E
F C^ ) = P (E) · P (F C^ |E) = (0.69) · 0 .17 = 0.1173.
Note that this is also P (E) − P (E
F ) = 0. 69 − 0 .5727.
(^) We can see that P (EC^
F ) = P (EC^ ) · P (F |EC^ ) = (0.31) · (0.58) = 0.1798.
Finally, P (EC^
F C^ ) = P (EC^ )·P (F C^ |EC^ ) = (0.31)·(0.42) = 0.1302.
(^) Thus, the probability of “not a good grade” (F C^ ) is simply found by adding the cases with F C^ in them. Thus the answer is 0.1173 + 0.1302 = 0.2475, i.e. 24.75%.
Problem 3 continued.
(^) First, let Sue sit anywhere. Even though, this feels like 10 choices, it is really just one choice due to circular arrangement!.
(^) Next, let Bob sit next two her. This can be done in 2 ways
(^) The rest of the people occupy the remaining 8 seats in P (8, 8) = 8! ways. Note that we no longer use circular arrangement, since our circle is now fixed with Bob and Sue at the head!
(^) By multiplication principle, the answer is:
n(E) n(S)
Problem 4.
(^) There are 40 different versions of an online homework. These are assigned randomly to the students. If a group of 3 classmates decide to work on the homework together, what is the probability that at least two of them receive the same version? Hint: It is easier to calculate the probability of this event not happening.
(^) Answer: The sample space consists of possible assignments of 3 of the 40 homeworks to our students, with repetition allowed! So, by multiplication principle n(S) = (40)(40)(40).
(^) As suggested we consider the event E that our students get three different homeworks first. Clearly n(E) = P (40, 3) = (40)(39)(38).
(^) So P (E) = (40)(39)(38) 403 = 14821600. Subtracting this from 1, we get the desired answer 0.074.
Problem 5.
A raffle held by Mundane Products has collected 900 entries. They are giving away 60 TVs and 30 stereos. John bought 14 entries for the raffle. (i) What is the probability that John wins exactly one TV as well as exactly one stereo? (ii) John just heard that the TVs are already decided and he has won exactly one of the 60 TVs. What is the probability that John will win at least one stereo in the remaining drawing?
(^) Answer: The sample space has n(S) = C(900, 14) sample points of choices of 14 out of 900 tickets.
Let E is the event of John winning 1 TV and 1 stereo and nothing more! Then he would have chosen one of the 60 TV tickets, (C(60, 1)), one of the 30 stereo tickets (C(30, 1)) and remaining 12 loser tickets (C(810, 12)).
Problem 5 continued.
(^) The desired probability is:
(60)(30)(810!) (12!)(798!)
(^) This has to be calculated by simplification and either typing in the simplified fraction carefully or by careful evaluation. It comes out 0.1165.
(^) For part (ii), take out all the TV tickets and now the sample space has n(S) = C(840, 13), since John still has 13 more tickets to choose.
(^) It would be easier to count the number of ways in which John does not win any stereos. That would be C(810, 13) when he chooses all loser tickets! This probability is C(810,13) C(840,13) and the final answer is 1^ −^
C(810,13) C(840,13) = 0.3789.
problem 7 continued.
(^) For part (i), the calculation is simple. Of the 310 managers, 95 got laid off, so the probability is estimated as 95 310 = 0.306. (^) For part (ii), we estimate probability of being a manager as 310 970 and the probability of being laid off as^
265
- The product of these numbers is 0.0873. The estimated probability of a manager being laid off from the given data is 97095 = 0.0979.
(^) Since the difference between the two answers is less than 0 .05, we estimate the probabilities as same, so the events are considered independent.
Problem 8.
(^) What is the probability that at least 2 people in a random group of 6 people have a birthday on the same day of the week?
(^) Answer: There are 7 different weekdays and let us consider the probability of the complementary event of the six people having distinct birth weekdays.
The probability is clearly P^ (7 76 , 6) = 0.0428. So, the probability of our event is 1 − 0 .0428 = 0.9572.
Problem 9.
A box of 200 printer cartridges is used to stock the shelves. 20 of the cartridges are defective. i) A customer buys 10 cartridges from the 200. What is the probability that all the ones he bought are good? ii) A second customer buys 10 more cartridges from the same shelf. What is the probability that he got all good ones too, assuming the first customer got all good ones?
(^) Answer: As before, we set the sample space as number of choices of 10 cartridges out of 200 or n(S) = C(200, 10).
Problem 9 continued.
(^) For part (i), the event E consists of choosing of 10 cartridges which have no defective ones, so n(E) = C(180, 10). So, P (E) = C C(180(200,,10)10) = 0.3398.
For part (ii), now we have only 190 cartridges with 20 defective. So, for the second customer, with similar calculation, we get the answer: C C(170(190,,10)10) = 0.3195.
Problem 11.
In an experiment, two events E and F are known to have probabilities 0.500 and 0.650 respectively, and it is suspected that the events are independent. However the probability of the event E
F cannot be determined theoretically. So the experiment is run 200 times and the event E
F
occurred 199 times. Based on this data, you would tend to think that the events E and F are/are not (choose one) independent, because the product of 0.500 and 0.650 is too large / too small / close to what it should be (choose one).
Answer. We estimate P (E
F ) by the formula P (E) + P (F ) − P (E
F ) = 0.5 + 0. 65 − 199 /200 = 0.155.
(^) Independence would require (0.5)(0.65) = 0.325 to be close to this estimated probability 0.155. Clearly, the product is too large and the event are not independent!
