EXAMPLE 3.5.5

At the entrance to a casino, there are two slot machines. Machine A is programmed so that in the long run it will produce a winner in 10% of the plays. Machine B is programmed so that in the long run it will produce a winner in 15% of the plays.

1. If we play each machine once, what is the probability that we will win on both plays?

2. If we play each machine once, what is the probability that we will lose on both plays?

3. If we play each machine once, what is the probability that we will win on at least one play?

SOLUTIONS

Let E be the event that we win when play Machine A; then P(E) = .1. Let F be the event that we win when we play machine B; the P(B) = .15.

1.

2.

3. Note that the event "Win at on at least one play" is the opposite (complement) of the event "lose on both plays." Thus we can use the complements rule and refer to the result of the previous problem:

P(win on at least one play) = 1 - P(lose on both plays) = 1 - .765 = .235.

Alternative solution

According to a formula from Unit 3 Module 5, P(E or F) = P(E) + P(F) - P(E and F)

= .1 + .15 - (.1)(.15) = .235