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:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → $\frac{5}{8}$
Step 1: Find the probability of non-selection for each person
- Probability that A is selected $P(A)$: $\frac{1}{4}$
- Probability that A is NOT selected $P(A')$: $1 - \frac{1}{4} = \frac{3}{4}$
- Probability that B is selected $P(B)$: $\frac{1}{6}$
- Probability that B is NOT selected $P(B')$: $1 - \frac{1}{6} = \frac{5}{6}$
Step 2: Calculate the combined probability
The probability that none of them is selected is the product of their individual non-selection probabilities:
$P(\text{None selected}) = P(A') \times P(B')$
$P(\text{None selected}) = \frac{3}{4} \times \frac{5}{6}$
Step 3: Simplify the result
$P(\text{None selected}) = \frac{3 \times 5}{4 \times 6} = \frac{15}{24}$
Dividing both numerator and denominator by 3:
$P(\text{None selected}) = \frac{5}{8}$
Final Answer: The probability that none of them is selected is $5/8$.