4 cards are drawn from a standard deck of 52 cards. The probability o…

The correct answer is Option (3) → \( \displaystyle\frac{({}^{13}C_{2})({}^{39}C_{2})}{{}^{52}C_{4}}\times 4 \)

Note: The given answer is as per the official NTA Answer Key. However, the answer given by NTA appears to be incorrect, because it counts some hands twice, and the correct probability is not among the options, as explained below.

Choosing exactly 2 cards of one particular suit, say hearts, and the other 2 from the remaining 39 cards:

    \( {}^{13}C_{2}\times{}^{39}C_{2} = 78\times741 = 57798 \) hands

Multiplying by 4 for the four suits gives 231192. But a hand such as 2 hearts + 2 spades has exactly 2 hearts and exactly 2 spades, so it is counted twice. The number of such hands is

    \( {}^{4}C_{2}\times({}^{13}C_{2})^2 = 6\times78^2 = 36504 \)

    Correct number of hands \( = 231192 - 36504 = 194688 \)
P \( = \displaystyle\frac{194688}{{}^{52}C_{4}} = \displaystyle\frac{194688}{270725} \)

Option (1) \( \displaystyle\frac{({}^{26}C_{2})({}^{26}C_{2})}{{}^{52}C_{4}} \) — Incorrect. It counts 2 red and 2 black cards, which is about colours, not suits.

Option (2) \( \displaystyle\frac{({}^{13}C_{2})({}^{39}C_{2})}{{}^{52}C_{4}}\times 3 \) — Incorrect. There are 4 suits, not 3.

Option (3) \( \displaystyle\frac{({}^{13}C_{2})({}^{39}C_{2})}{{}^{52}C_{4}}\times 4 \) — The answer given by NTA, but it seems to be incorrect. It counts every hand with 2 cards of one suit and 2 of another suit twice, so it is larger than the correct value.

Option (4) \( \displaystyle\frac{({}^{26}C_{2})({}^{26}C_{2})}{{}^{52}C_{4}}\times 3 \) — Incorrect. It is about colours, not suits.

Advice for the exam: In the real exam, mark the answer that is mathematically correct. If NTA’s provisional answer key marks a different option, challenge it during the answer-key challenge window, giving proper reasoning. NTA does revise its key when a challenge is well justified, so a sound challenge can raise your marks.