Past records have led a bank to believe that the probability

Past records have led a bank to believe that the probability that a customer opens a savings

account is 0.6, the probability that a customer opens a checking account is 0.7, and the

probability that a customer opens a savings and a checking account is 0.52.

(a) What is the probability that a customer will open at least one of the two accounts?

(b) What is the probability that a customer will open exactly one of the two accounts?

Let S = customer opens a savings account

C = customer opens a checking account

(d) Compute the following probabilities:

let A denote the event of openina savings account

therefore P(A) = 0.6--given

let B denote the event of opening a checking acount

therefore P(B) = 0.7 --given

and P(A and B) = 0.52

a) the probability that a customer will open at least one of the two accounts

is P( A or B) = P(A) + P(B) - P(A and B) = 0.6 + 0.7 - 0.52 = 0.78

b) the probability that a customer will open exactly one of the two accounts

is P(A or B) - P(A and B) = 0.78 - 0.52 = 0.26

c) the probability that a customer will open none of the two accounts

is 1-P( A or B ) = 1 - 0.78 = 0.22

d)
1) P(S and C\') = P(S) - P( S and C)
= 0.6 - 0.52 = 0.08

2) P( S\' or C) = 1 - P(S)
= 1-0.6 = 0.4

3) P(S\'/C) = P(s\' and C)/P(C)

P(S\' and C) = P(C) - P(S and C)
= 0.7 - 0.52 = 0.18

P(s\' and C)/P(C) = 0.18/0.7 = 0.2571

e) Two events are mutually exclusive if they cannot occur at the same time.

ie., P(S and C) should be zero

but here P(S And C) = 0.52 which is not zero

so the events are not mutually exclusive