Question

A money lending company employs analysts to manage loan accounts for its customers. In an analyst’s portfolio, a quarter of the female customer accounts are in arrears whilst three fifth of the male customer accounts are up to date. Company records show that 34% of this analyst’s customers are females and the rest are male. If a randomly selected account from this analyst’s portfolio is in arrears, what is the probability (to four decimal places) that the account holder is female?

Answer 1

$P(Female) = 0.34 \\ P(Male)= 1 -0.34 = 0.66 \\ P(Arrear|Female) = \frac{1}{4}= 0.25 \\ P(Arrear|Male) = \frac{2}{5}=0.4 \\ P(Female|Arrear) = \frac{P(Female) \times P(Arrear|Female)}{P(Female) \times P(Arrear|Female) +P(Male) \times P(Arrear|Male)} \\ = \frac{0.34 \times 0.25}{(0.34 \times 0.25) +(0.66 \times 0.4)} \\ = 0.2436$

If a randomly selected account from this analyst’s portfolio is in arrears, the probability that the account holder is female is 0.2436.