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Two candidates, X and Y appeared in an interview for two vacancies for the same post. The probabilities of their selection are 1/4 and 1/7, respectively. What is the probability that one of them will be selected?

Solution

✅ Correct Option: 4

Given:

Probability of X being selected = 14\frac{1}{4}

Probability of Y being selected = 17\frac{1}{7}

The question asks for the probability that exactly one of them will be selected.


The probability that X is not selected:

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

P(X not selected)=34P(\text{X not selected}) = \frac{3}{4}

The probability that Y is not selected:

P(Y not selected)=1−17P(\text{Y not selected}) = 1 - \frac{1}{7}

P(Y not selected)=67P(\text{Y not selected}) = \frac{6}{7}


Exactly one candidate is selected when either:

  • X is selected and Y is not selected, OR
  • Y is selected and X is not selected

Probability that X is selected and Y is not selected:

P(X selected and Y not selected)=14×67P(\text{X selected and Y not selected}) = \frac{1}{4} \times \frac{6}{7}

P(X selected and Y not selected)=628P(\text{X selected and Y not selected}) = \frac{6}{28}


Probability that Y is selected and X is not selected:

P(Y selected and X not selected)=17×34P(\text{Y selected and X not selected}) = \frac{1}{7} \times \frac{3}{4}

P(Y selected and X not selected)=328P(\text{Y selected and X not selected}) = \frac{3}{28}


Probability that exactly one is selected:

P(exactly one selected)=628+328P(\text{exactly one selected}) = \frac{6}{28} + \frac{3}{28}

P(exactly one selected)=928P(\text{exactly one selected}) = \frac{9}{28}

Therefore, the probability that one of them will be selected is 928\frac{9}{28}.

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