From a standard deck of 52 playing cards, three cards are drawn at random. What is the probability that exactly one card is an ace, one card is a queen and one card is a jack?

Introduction / Context:
This problem tests basic probability with combinations on a standard deck of 52 playing cards. We are interested in the probability that in a draw of three cards, we get one ace, one queen and one jack, with order not specified but all three card types required exactly once.

Given Data / Assumptions:

• There are 52 distinct cards in a standard deck.
• There are 4 aces, 4 queens and 4 jacks in the deck.
• Three cards are drawn at random without replacement.
• Any order of drawing is acceptable as long as the final set contains exactly one ace, one queen and one jack.

Concept / Approach:
We use combinations to count the total possible sets of 3 cards and the favourable sets. Since order does not matter for combinations, but the sequence of drawing can differ, we must be careful to count correctly. The probability is equal to favourable outcomes divided by total outcomes.

Step-by-Step Solution:
Total ways to choose 3 cards from 52: 52C3.52C3 = 52 * 51 * 50 / (3 * 2 * 1) = 22100.Ways to choose 1 ace from 4: 4C1 = 4.Ways to choose 1 queen from 4: 4C1 = 4.Ways to choose 1 jack from 4: 4C1 = 4.Since the draw is of 3 distinct cards and each chosen once, the number of favourable 3 card sets is 4 * 4 * 4 = 64.However, each 3 card set can occur in 3! = 6 different orders when drawing, so the total favourable ordered outcomes is 64 * 6 = 384.The probability is therefore 384 / 22100.Simplify: 384 / 22100 = 96 / 5525.

Verification / Alternative check:
Another way is to think in terms of choosing card types and suits directly. On each draw we want one from the ace group, one from the queen group and one from the jack group. Counting distinct combinations of ranks and suits leads to the same count of 64 unordered sets and therefore the same probability 96 / 5525 after accounting for order. This confirms our computation.

Why Other Options Are Wrong:
80/5525, 64/5525 and 48/5525 all represent smaller probabilities than the correct value. They would correspond to incorrectly counting fewer favourable outcomes, for example by missing some suit possibilities or by omitting the effect of different drawing orders. None of these fractions matches the correct ratio of 384 favourable outcomes out of 22100 total outcomes.

Common Pitfalls:
A common mistake is to compute 4C1 * 4C1 * 4C1 = 64 and then divide directly by 52C3 without considering whether order matters. In this problem, we explicitly count unordered sets and then relate them to the total number of unordered sets, giving 64 / 52C3. After simplification that becomes 96 / 5525. Another frequent error is to treat the events as independent with replacement, which is not correct here because cards are not replaced once drawn.

Final Answer:
The required probability is 96/5525.