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Probability

Section: Mathematics

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Question : 35 of 92

Marks: +1, -0

70\%

of the total employees of a factory are men. Among the employees of that factory,

30\%

15\%

of women are technical assistants. If an employee chosen at random is found to be a technical assistant, then the probability that this employee is a man is

[AP EAPCET 21 May 2025 S1]

$\frac{9}{23}$
$\frac{3}{17}$
$\frac{14}{17}$
$\frac{14}{23}$

Solution:

Let

E_1=

Employee be a man

E_2=

Employee be a woman

E=

Employee be a technical assistants

Given

n(E_1) = \; \frac{70}{100} = \; \frac{7}{10}, P(E_2) = \; \frac{3}{10}

\;P\left(\frac{E}{E_1}\right) =\; \frac{3}{10}

\;P\left(\frac{E}{E_2}\right) =\; \frac{15}{100}

P\left(\frac{E_1}{E}\right) = \frac{P(E_1) P\left(\frac{E}{E_1}\right)}{P(E_1) P\left(\frac{E}{E_1}\right) + P(E_2) P\left(\frac{E}{E_2}\right)}

\; = \frac{\; \frac{7}{10} \times \frac{3}{10}}{\frac{7}{10} \times \; \frac{3}{10} + \; \frac{3}{10} \times \; \frac{15}{100}}

\; = \frac{210}{210+45} = \; \frac{210}{255}

P\left(\frac{E_1}{E}\right) = \frac{42}{51} = \; \frac{14}{17}

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