[2/7] The probability A getting a job is 1/5 and that of B is 1/7 . What is the probability that only one

The probability A getting a job is 1/5 and that of B is 1/7 . What is the probability that only one of them gets a job?

11/35

12/35

2/7

1/7

Explanation:

Let A = Event that A gets a job and B = Event that B gets a job

Given that $P\left(A\right)$ = 1/5 and $ P\left(B\right)$ = 1/7

Probability that only one of them gets a job

$= \text{P}\left[\left(A \cap \bar{\text{B}}\right)\cup \left(\text{B} \cap \bar{\text{A}}\right)\right]$[ Reference :Algebra of Events]

$= \text{P}\left(A \cap \bar{\text{B}}\right)+ \text{P}\left(\text{B} \cap \bar{\text{A}}\right)$[ Reference :Mutually Exclusive Events and Addition Theorem of Probability]

$= \text{P(A)P(}\bar{\text{B}}) + \text{P(B)P(}\bar{\text{A}})$[ Here A and B are Independent Events and refer theorem on independent events]

$= \text{P(A)}\left[1 - \text{P(B)}\right] + \text{P(B)}\left[1 - \text{P(A)}\right]$

$= \dfrac{1}{5}\left(1 - \dfrac{1}{7}\right) + \dfrac{1}{7}\left(1 - \dfrac{1}{5}\right) = \dfrac{1}{5} \times \dfrac{6}{7} + \dfrac{1}{7}\times \dfrac{4}{5} = \dfrac{6}{35} + \dfrac{4}{35} = \dfrac{10}{35} = \dfrac{2}{7}$

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