14 April, 08:57

# 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?

+3
1. 14 April, 09:22
0
96%

Step-by-step 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