A pack contains 6 cards numbered 1, 2, 3, 4, 5 and 6. Four cards are drawn one by one at random without replacement. What is the probability that the card numbered 1 is drawn in the selected four cards?

Correct option is A

Given:

Total cards = 6 (numbered 1, 2, 3, 4, 5, 6)
4 cards are drawn randomly without replacement.
Find the probability that card numbered 1 is drawn.

Formula:

$$\text{Probability} = \frac{\text{Favorable outcomes}}{\text{Total outcomes}}$$

​Where:
Favorable outcomes = Number of ways to select 3 cards from the remaining 5 cards (because card 1 is already selected).
Total outcomes = Number of ways to select any 4 cards from 6 cards.

Combinations formula: $$^nC_r = \frac{n!}{r!(n-r)!}$$

​Solution:

$$\text{Favorable outcomes} = {^5C_3} = \frac{5!}{3!2!} = \frac{5 \times 4}{2 \times 1} = 10 \\\\\ \\\text{Total outcomes} = {^6C_4} = \frac{6!}{4!2!} = \frac{6 \times 5}{2 \times 1} = 15 \\[10pt]\\\ \\\text{Thus, Probability} = \frac{10}{15} = \frac{2}{3} \\[10pt]\text{Option (a):} \quad \frac{4}{6} = \frac{2}{3} \quad \text{(matches)} \\[10pt]$$

​Final Answer: (a) 4/6