Question

Among users of automated teller machines​ (ATMs), 94​% use ATMs to withdraw cash and 28​% use them to check their account balance. Suppose that 95​% use ATMs to either withdraw cash or check their account balance​ (or both). Given a woman who uses an ATM to check her account​ balance, what the probability that she also uses an ATM to get​ cash?

Answer 1

96%

Explanation:

Conditional probability is defined as:

P(A|B) = P(A∩B) / P(B)

Or, in English:

Probability that A occurs, given that B has occurred = Probability that both A and B occur / Probability that B occurs

We want to find the probability that a woman uses an ATM to get cash, given that she uses an ATM to check her balance.

P(withdraws cash | checks account)

Using the definition of condition probability, this equals:

P = P(withdraws cash AND checks account) / P(checks account)

We know that P(checks account) = 0.28.

But we don't know what P(withdraws cash AND checks account) is. To find that, we need to use the definition of P(A∪B):

P(A∪B) = P(A) + P(B) − P(A∩B)

This says that the probability of A or B occurring (or both) is the probability of A occurring plus the probability of B occurring minus the probability of both A and B occurring.

P(withdraws cash OR checks account) = P(withdraws cash) + P(checks account) − P(withdraws cash AND checks account)

0.95 = 0.94 + 0.28 − P(withdraws cash AND checks account)

P(withdraws cash AND checks account) = 0.27

Therefore:

P = 0.27 / 0.28

P ≈ 0.96