A bag contains 17 tickets numbered from 1 to 17. A ticket is drawn, a…

The correct answer is Option (3) → \( \displaystyle\frac{25}{34} \)

Note: The given answer is as per the official NTA Answer Key. However, the wording “both the tickets show at least one even number” does not make clear whether at least one of the two tickets or each of the two tickets must show an even number, and with the second reading Option (1) could also be a possible answer, as explained below.

Among 1 to 17 there are 8 even and 9 odd numbers, and the tickets are drawn without replacement.

At least one of the two tickets even: use the complement, both tickets odd.

    \( \displaystyle P(\text{both odd})=\frac{9}{17}\cdot\frac{8}{16} \)

    \( \displaystyle =\frac{9}{34} \)

    \( \displaystyle P(\text{at least one even})=1-\frac{9}{34}=\frac{25}{34} \)

Each of the two tickets even:

    \( \displaystyle P(\text{both even})=\frac{8}{17}\cdot\frac{7}{16} \)

    \( \displaystyle =\frac{7}{34} \)

Option (1) \( \displaystyle\frac{7}{34} \) — Should also be correct, though not the answer given by NTA. It is the probability that each of the two tickets shows an even number, the second reading of the wording.

Option (2) \( \displaystyle\frac{27}{34} \) — Incorrect. Neither reading gives this value; the two readings give \( \frac{25}{34} \) and \( \frac{7}{34} \).

Option (3) \( \displaystyle\frac{25}{34} \) — Correct. It is the probability that at least one of the two tickets shows an even number, 1 minus the chance that both are odd.

Option (4) \( \displaystyle\frac{9}{34} \) — Incorrect. It is the probability that both tickets are odd, that is, that no ticket shows an even number.