Introduction to Probability and Statistics for Engineers and Scientists

3.6Conditional Probability 69

SOLUTION Since the transistor did not immediately fail, we know that it is not one of the
5 defectives and so the desired probability is:

P{acceptable|not defective}

=

P{acceptable, not defective}

P{not defective}

=

P{acceptable}

P{not defective}

where the last equality follows since the transistor will be both acceptable and not defective
if it is acceptable. Hence, assuming that each of the 40 transistors is equally likely to be
chosen, we obtain that

P{acceptable|not defective}=

25/40

35/40

=5/7

It should be noted that we could also have derived this probability by working directly
with the reduced sample space. That is, since we know that the chosen transistor is not
defective, the problem reduces to computing the probability that a transistor, chosen
at random from a bin containing 25 acceptable and 10 partially defective transistors, is
acceptable. This is clearly equal to^2535. ■

EXAMPLE 3.6b The organization that Jones works for is running a father–son dinner for
those employees having at least one son. Each of these employees is invited to attend along
with his youngest son. If Jones is known to have two children, what is the conditional
probability that they are both boys given that he is invited to the dinner? Assume that the
sample spaceSis given byS={(b,b), (b,g), (g,b), (g,g)}and all outcomes are equally
likely [(b,g) means, for instance, that the younger child is a boy and the older child is
a girl].

SOLUTION The knowledge that Jones has been invited to the dinner is equivalent to know-
ing that he has at least one son. Hence, lettingBdenote the event that both children are
boys, andAthe event that at least one of them is a boy, we have that the desired probability
P(B|A) is given by

P(B|A)=

P(BA)

P(A)

=

P({(b,b)})

P({(b,b), (b,g), (g,b)})

=

1

4

3

4

=

1

3

Many readers incorrectly reason that the conditional probability of two boys given at least
one is^12 , as opposed to the correct^13 , since they reason that the Jones child not attending