Question

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

 

a) The probability they have a current account or a savings account.

 

b) The probability they have a current account and a savings account.

 

c) The probability they have a current account but not a savings account.

 

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

Answer 1

Solution:

Denote the persons having Savings account by S

and persons having Current account by C.

$P(S)=60\%=0.6 \\P(C)=90\%=0.9$

(a):

Every bank customer has one or both of these. It means

The probability they have a current account or a savings account$=P(S\cup C)=100\%=1$

(b):

b) The probability they have a current account and a savings account$=P(S\cap C)$

$=P(S)+P(C)-P(S\cup C) \\=0.6+0.9-1 \\=0.5 \\=50\%$

(c): The probability they have a current account but not a savings account$=P(C \cap S')$

$=P(C)-P(C \cap S) \\=0.9-0.5 \\=0.4 \\=40\%$

(d):

The probability they do not have a savings account, given that they have a current account$=P(S'|C) \\=\dfrac{P(S'\cap C)}{P(C)} \\=\dfrac{0.4}{0.9} \\=\dfrac49$