Question

The probability a getting a job is 1/5 and that of b 1/7.what is rhe probability that only one of them get a job?

Answer 1

if is
sum of a's success b failure +a"s failure b success
$( \frac{1}{5} )(1 - \frac{1}{7} ) + ( \frac{1}{7} )(1 - \frac{1}{5} ) \\ = ( \frac{1}{5} )( \frac{6}{7} ) + ( \frac{1}{7} )( \frac{4}{5} ) \\ = \frac{6}{35} + \frac{4}{35} \\ = \frac{10}{35} \\ = \frac{2}{7}$
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Answer 2

Given :  The probability a getting a job is 1/5 and that of b 1/7

To find :  probability that only one of them get a job

Solution:

Probability of getting a job = 1/5

Probability of a not getting job = 1 - 1/5 = 4/5

Probability of getting a job = 1/7

Probability of a not getting job = 1 - 1/7 = 6/7

probability that only one of them get a job

= a got Job & b does not get job + b get job a does not get

= (1/5)(6/7) + (1/7)(4/5)

= 6/35 + 4/35

= 10/35

= 2/7

Another Method

Probbaility = 1 - both get job - none get job

= 1 - (1/5)(1/7) - (4/5)(6/7)

= 1 - 1/35 - 24/35

= (35 - 1 - 24)/35

= 10/35

= 2/7

2/7 is the probability that only one of them get  job

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