The probability distribution of the number of aces when two cards are drawn without replacement from a pack of 52 cards is :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) →
| \(x_i\) | 0 | 1 | 2 |
| \(p_i\) | \(\displaystyle\frac{188}{221}\) | \(\displaystyle\frac{32}{221}\) | \(\displaystyle\frac{1}{221}\) |
A pack has 4 aces and 48 other cards. Two cards drawn without replacement can be chosen in
\( \displaystyle {}^{52}C_{2} = \frac{52\times 51}{2} = 1326 \) ways.
No ace (both cards from the 48):
\( \displaystyle P(X = 0) = \frac{{}^{48}C_{2}}{1326} = \frac{1128}{1326} = \frac{188}{221} \)
One ace and one other card:
\( \displaystyle P(X = 1) = \frac{4\times 48}{1326} = \frac{192}{1326} = \frac{32}{221} \)
Two aces:
\( \displaystyle P(X = 2) = \frac{{}^{4}C_{2}}{1326} = \frac{6}{1326} = \frac{1}{221} \)
Check: 188 + 32 + 1 = 221, so the probabilities add up to 1.
Why the other options are incorrect
Option (1) \(p_i\) = 144/221, 24/221, 1/221 for \(x_i\) = 0, 1, 2 — Incorrect. These values add up to 169/221, not 1.
Option (2) \(p_i\) = 144/169, 24/144, 1/144 for \(x_i\) = 0, 1, 2 — Incorrect. 144/169 is P(X = 0) with replacement, and the three values do not add up to 1.
Option (4) \(p_i\) = 188/221, 24/221, 1/221 for \(x_i\) = 0, 1, 2 — Incorrect. P(X = 1) is 32/221; these values add up to 213/221, not 1.