Probability of Drawing Two Aces – UGC NET Question

There are 4 aces in a standard 52-card deck. The probability that the first card is an ace is 4/52, since any of the 4 aces could appear among the 52 cards. After one ace has been drawn, 3 aces remain in a reduced deck of 51 cards, so the probability that the second card is also an ace is 3/51. Because the draws are without replacement and the events are dependent, we multiply the probabilities: (4/52) × (3/51) = 12/2652, which simplifies to 1/221. Therefore, the correct probability that both drawn cards are aces is 1/221.

Option A:
Option A (1/13) usually comes from considering only the probability that a single card is an ace (4/52 = 1/13) and ignoring that we need two consecutive aces. It does not incorporate the second dependent draw, so it does not represent the joint probability of both events occurring. Because it treats the problem as if only one ace were required, it significantly overestimates the true probability. Hence, 1/13 is not correct for “both cards are aces.”

Option B:
Option B (1/169) typically arises from incorrectly assuming that the two draws are independent with replacement and computing (1/13) × (1/13). However, in this question the cards are drawn without replacement, which means the second probability changes after the first card is drawn. Using 1/169 ignores the reduced deck on the second draw and does not reflect the correct dependent-event structure. Therefore, it underestimates the correct probability.

Option C:
Option C (2/221) looks close to the correct answer but is exactly double 1/221, and there is no step in the correct derivation that would justify multiplying the final probability by 2. This kind of error often happens when students miscount favourable outcomes or mistakenly think order does not matter when they have already accounted for it. Since the exact product (4/52) × (3/51) simplifies to 1/221, any multiple of that value, including 2/221, is incorrect.

Option D:
Option D (1/221) matches the precise result from multiplying the probability of the first ace and the second ace without replacement: (4/52) × (3/51). The numerator 12 counts the ordered ways to pick two aces in succession, and the denominator 2652 counts all ordered pairs of cards, which simplifies to 1/221. This value correctly reflects both the limited number of aces and the shrinking deck size after the first draw. Hence, it is the only option fully consistent with the rules of probability for dependent events.