Notes on Problems for Quiz 3 (10/10/97) from Schaum:

4.35, 4.37, 4.40, 4.43, 4.49, 4.54, 4.60, 4.61, 4.62, 4.64

Problem 4.35:

(1) This example of "conditional probability" really can be viewed without the basic definition of conditional probability on page 54 of the text. It can be regarded as a probability problem where the equiprobable sample space is the set of all selections of four cards from a deck of cards with three spades removed. Thus, there are 10 spades and 13 of each of the three other suits. The probability of exactly zero or one spade in the four cards selected from this deck of 49 cards is

(C(39,4)C(10,0)+C(39,3)C(10,1))/C(49,4).

The answer, about 0.18, is 1 minus this number.

Problem 4.37:

(1) For (i), South has a total of 13 cards (and exactly one ace). There are C(39,13) ways that North can be delt a hand. If we assume that North has the other three aces, then imagine giving them to North first.There are then C(36,10) ways of dealing the rest of the cards to North. Thus, the probability of North having the other three aces is C(36,10)/C(39,3). For (ii), the two events, West has the other three hearts and East has the other three hearts are disjoint. Thus the required probability is the sum of the two events which is 2 times the probability that West has the other three hearts. Thus the required probability is 2xC(23,13)/C(26,13).

Problem 4.40:

(1) Let BH be the set of people with Brown Hair and BE the set of people with Brown Eyes. Using the information given, we compute the probabilities of the basic (contiguous) regions of the Venn Diagram for BH intersect BE as follows:

We can, using this diagram, answer (i), (ii), and (iii) easily. For (i) we get .15/.40, for (ii) we get .10/.25, and for (iii) we get .50 directly from the Venn diagram.

Problem 4.43:

(1) Here is what we are given in the problem P(Boy)=.4, P(Girl)=.6, P(Math | Girl)=.1, P(Math | Boy)=.25. We are asked to find P(Girl | Math). From Bayes's Theorem

P(Girl | Math) equals P(Math | Girl) P(Girl) / (P(Math | Girl) P(Girl) + P(Math | Boy) P(Boy)).

Thus P(Girl | Math) = 0.1 x 0.6 / (0.1 x 0.6 + 0.25 x 0.4) = 6 / 16 = 3 / 8.

Problem 4.49:

(1) The tree-diagram was discussed in class and is shown on page 72 of the text. You can also use conditional probabilities. For (i), we get P(3 or 6 appears)P(Marble transferred is red | 3 or 6 appears)P(Marble chosen is red | Marble transferred is red and 3 or 6 appears) = 1/3 x 1/3 x 6/9 = 2 / 27. To this we must add P(1,2,4,5 appears)P(Marble transferred is red | 1,2,4,5 appears)P(Marble chosen is red | 1,2,4,5 appears and marble transferred is red) = 2/3 x 5/8 x 1/2 = 10 / 48 = 5 /24. The sum of 2 / 27+ 5 / 24 = 61 / 216. For (ii) a similar calculation can be done. In both cases the tree diagram is easier. The tree is shown on page 72 of the text.

Problem 4.54:

(1) This problem assumes that it is known that exactly 2 tubes are defective. A more common assumption is that of (2) below. Assuming exactly 2 tubes are known to be defective, then the probability of the test sequence starting off DD ... is 2/5 x 1/4 = 2 / 20 = 1 / 10. The process can be stopped on the third test if we get any of the sequences DND, NDD, or NNN. The latter sequence, NNN, results because we know that there are exactly two defective tubes, so we know that if NNN appears, then tubes 4 and 5 are defective. These probabilities are 2/5 x 3/4 x 1/3 + 3/5 x 2/4 x 1/3 + 3/5 x 2/4 x 1/3= 1/10 + 1/10 + 1/10 = 3/10.

(2) A more realistic problem is that five tubes are tested one after another. If two defective tubes are found then the whole batch is discarded. What is the probability that the whole batch is discarded on the third test?

Problem 4.60:

(1) For each p>0 and q>0, we have a sample space S(p,q) which is the set of all pairs (x,y) where x is a sequence of p symbols H or M and y is a sequence of q symbols H or M. The sequence x is the "hit - miss" sequence for A firing p times and y is the "hit - miss" sequence for B firing q times. In (i) we compute 1 - P(no hits). P(no hits) is (3/4)(3/4)(2/3)(2/3) = 1/4. The required probability is 3/4. For (ii) the samples where there is one H consists of (H,M) and (M,H) with respective probabilities (1/4)((2/3) and (3/4)(1/3) for a total of 5/12. The event where A hits the target and there is only one hit consists of the single sample point (H,M) which has probability 2/12. The required conditional probability is (2/12)/(5/12) = 2/5. For (iii), we need to consider the general situation where we have (x,y) with x consisting of a sequence of length 2 and y a sequence of length t, t unknown, but satisfying (3/4)^2 (2/3)^t < 0.10. The notation v^t means raise v to the power t. Thus, (2/3)^t < 16/90 or t > ln(16/90) / ln(2/3) = 4.25. Since t is an integer, we have t greater than or equal to 5.

Problem 4.61:

(1) For part (i) we use the fact that 2/3 = 1/2 + P(B) - (1/2)P(B). Here we use Theorem 3.5 on page 41 together with the fact that A and B are independent (Definition page 57). Solving for P(B) gives P(B)=1/3. For part (ii), P(A|B) = P(A) = 1/2 from independence. For part (iii), we showed in class that if X and Y are independent events in a sample space S then S-X and Y are independent (S-X is the complement of X). Thus P(S-B|A) = P(S-B) = 1 - P(B) = 2/3.

Problem 4.62:

(1) The key idea here is that we showed in class that if X, Y, and Z are indendent events in a sample space S then S-X, Y, and Z are independent. In words, if three events are independent then the three events obtained by complementing any one of them are also independent. If you believe that, then use it again to get S-X, S-Y, Z are independent of S-X, Y, and S-Z are independent. Use it one more time to get S-X, S-Y, and S-Z are independent. In other words, once you now you can complement one of the events and retain independence, you can get any combination of complemented events independent.

Problem 4.64:

(1) Make up a table:

The desired probability is the sum of the entries common to the row labled W and the column labeled W.