Two cards are drawn simultaneously at random from a well shuffled pack of 52 Cards. Let X be the random variable which denotes number of kings in the draw. Then the probability distribution of X is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) →
| X | 0 | 1 | 2 |
| P(X) | \(\displaystyle\frac{^{48}C_{2}}{^{52}C_{2}}\) | \(\displaystyle\frac{^{4}C_{1} \times {}^{48}C_{1}}{^{52}C_{2}}\) | \(\displaystyle\frac{^{4}C_{2}}{^{52}C_{2}}\) |
The pack has 4 kings and 48 other cards. The two cards can be drawn in \(^{52}C_{2}\) ways, and X can take the values 0, 1 and 2. For each value, the kings and non-kings are chosen separately:
X = 0: both cards from the 48 non-kings, in \(^{48}C_{2}\) ways
X = 1: one king and one non-king, in \(^{4}C_{1} \times {}^{48}C_{1}\) ways
X = 2: both cards from the 4 kings, in \(^{4}C_{2}\) ways
Dividing each by \(^{52}C_{2}\):
\( \displaystyle P(X = 0) = \frac{^{48}C_{2}}{^{52}C_{2}} \)
\( \displaystyle P(X = 1) = \frac{^{4}C_{1} \times {}^{48}C_{1}}{^{52}C_{2}} \)
\( \displaystyle P(X = 2) = \frac{^{4}C_{2}}{^{52}C_{2}} \)
As a check, \(^{48}C_{2} + {}^{4}C_{1} \times {}^{48}C_{1} + {}^{4}C_{2} = 1128 + 192 + 6 = 1326 = {}^{52}C_{2}\), so the probabilities add up to 1.