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:

$\frac{5}{8}$

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 $5/8$.