Answer to Question #256178 in Statistics and Probability for Mothichand

Question

A bank provides two different kind of bank account: current accounts and savings accounts. Every bank customer has one or both of these. 90% of bank customers have current accounts and 60% of bank customers have savings accounts. If a customer is chosen uniformly at random from the set of all bank customers, calculate the following probabilities, showing your working in each case:

Answer 1

Let $C$ be the event that a customer has a current account and $S$ be the event that a customer has a savings account.

Case 1

Since 90% of the customers have a current account, it implies that the remaining 10% of the customers have both a savings and current account because every customer has one or both accounts.

Case 2

Since 60% of the customers have a savings account, it implies that the remaining 40% have both a current account and a savings account because every customer has one or both accounts.

Since we do not have the exact number of customers in this bank, we are going to have a total of 200% customers. From Case 1 and Case 2, we can see that, 10%+40%=50% have both savings and current accounts.

This will help us compute the required probabilities as follows.

Define the probabilities

$p(C)=90\%/200\%=9/20$

$p(S)=60\%/200\%=3/10$

$p(C\cap S)=50\%/200\%=1/4$

a.

$p(C\cup S)$ is obtained using the addition formula of probability given as,

$p(C\cup S)=p(C)+p(S)-p(C\cap S)=9/20+3/10-1/4=10/20=1/2$

Therefore, the probability that they have a current account or a savings account is 1/2.

b.

$p(C\cap S)$ as mentioned above is given as,

$p(C\cap S)=(10\%+40\%)/200\%=50\%/200\%=1/4$

The probability that they have both a savings and current account is 1/4.

c.

 The probability they have a current account but not a savings account is given as,

$p(C\cap S')=p(C)-p(C\cap S)=9/20-1/4=1/5$

Therefore, probability that they a current account but not a savings account is 1/5.

d.

The probability they do not have a savings account, given that they have a current account is given as,

$p(S'|C)=p(C\cap S')/p(C)=(1/5)/(9/20)=20/45=4/9.$

Thus, the probability that they do not have a savings account given that they have a current account is 4/9.