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Two persons A and B appear in an interview for two vacancies. If the probabilities of their selections are 1/4 and 1/6 respectively, then the probability that none of them is selected shall be:

Solution

✅ Correct Option: 4

Two persons A and B appear in an interview for two vacancies.

Probability A is selected =14= \frac{1}{4}

Probability B is selected =16= \frac{1}{6}


The probability that A is not selected:

P(A not selected)=1−14P(\text{A not selected}) = 1 - \frac{1}{4}

=34= \frac{3}{4}


The probability that B is not selected:

P(B not selected)=1−16P(\text{B not selected}) = 1 - \frac{1}{6}

=56= \frac{5}{6}


Since A's selection and B's selection are independent events, the probability that both are not selected:

P(None selected)=P(A not selected)×P(B not selected)P(\text{None selected}) = P(\text{A not selected}) \times P(\text{B not selected})

=34×56= \frac{3}{4} \times \frac{5}{6}

=1524= \frac{15}{24}

=58= \frac{5}{8}

Therefore, the probability that none of them is selected =58= \frac{5}{8}

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