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A bank has studied its chequing accounts and found that 90% of all chequing accounts have been open for at least one year, the remaining percentage of chequing accounts have been open for less than a year. The bank also determined that for all chequing accounts that have been open for less than one year, the percentage of cheques returned due to insufficient funds is 4%. For chequing accounts that have been open for at least one year, only 1% of cheques were returned due to insufficient funds.

(a) What is the probability that a cheque processed by this bank will be returned due to insufficient funds?

(b) If a cheque is returned due to insufficient funds, what is the probability that it came from a bank account that has been open for more than one year?

Let P(a) = 0.01

since open for at least one year, only 1% of cheques were returned due to insufficient funds.

Let P(b) = 0.04

since open for less than one year, the percentage of cheques returned due to insufficient funds is 4%.

(a)

I added P(a) = 0.01 and Let P(b) = 0.04 to get 0.05

(b)

P(A ∩ B) = P(a)*P(b)

P(A ∩ B) = 0.01 * 0.9

P(A ∩ B) = 0.009

I got both answers wrong. What is the correct answer and solution?

1 Answers 1

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a) By the law of the total probability:

$$P(\text{return})=P(\text{return}|\text{less than 1 year)}P(\text{less than 1 year)}+P(\text{return}|\text{more than 1 year)}P(\text{more than 1 year)}\\ =0.04\cdot0.1+0.01\cdot0.9=0.013$$

b) By Bayes' law:

$$P(\text{more than 1 year}|\text{return})= \frac{P(\text{return}|\text{more than 1 year})P(\text{more than 1 year})}{P(\text{return})}\\ =\frac{0.01\cdot 0.9}{0.013}$$