CHAPTER 2 - Conditional Probability

Section 2.1, page 55

Problem 9

(a) x =  P(both red| A was selected) = P(both red & A selected)/P(A selected)

The number of ways to select cards is C(5,2), and the number of ways to select A and another card is C(4,1). The number of ways to select A and both red is C(3,1). So we have

>	x:=(binomial(3,1)/binomial(5,2))/(binomial(4,1)/binomial(5,2));

(or more simply put, if you know you already have A, then 3 of the remaining 4 cards are red).

(b) y = P(both red|at least one is red)= P(both red [at least one is red is redundant])/P(at least one is red) -- but since all but one of the cards is red, P(at least one is red) = 1. Therefore:

>	y:=binomial(4,2)/binomial(5,2);

Section 2.2, page 64

Problem 10

x=P(at least 3 blue | at least one blue) = P(at least 3)/P(at least one).

Now, y=P(at least 3)=P(3) + P(4) + P(5), and

>	y:=sum(binomial(5,k)*(1/4)^k*(3/4)^(5-k),k=3..5);

And z=P(at least one) = 1 - P(none), so

>	z:=1-(3/4)^5;

>

x:=y/z;

>

evalf(%);

So given that at least one child is blue-eyed, the probability is 13.6% that at least three of them are.

Section 2.3, page 77

Problem 10

(a) We want x = P(proper | 4 of 5 good). We know P(proper)=0.9, and P(4 of 5 good|proper) is C(5,1)*. We also need P(improper)=0.1 and P(4 of 5 good | improper) is C(5,1)*. So we can compute:

>	x:=0.9*binomial(5,1)*(1/2)^5/(0.9*binomial(5,1)*(1/2)^5+0.1*binomial(5,1)*(1/4)^4*(3/4));

So the chances are excellent that the machine was properly adjusted.