Two cards are drawn successively with replacement from a well shuffled pack of 52 playing cards. Which of the following is correct probability distribution of number of Queens(X) as \( \begin{pmatrix}X\\P(X)\end{pmatrix} \)?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \begin{pmatrix}0 & 1 & 2\\\displaystyle\frac{144}{169} & \displaystyle\frac{24}{169} & \displaystyle\frac{1}{169}\end{pmatrix} \)
The cards are drawn with replacement, so the two draws are independent and X follows a binomial distribution with n = 2.
\( \displaystyle p=P(\text{queen})=\frac{4}{52}=\frac{1}{13} \)
\( \displaystyle q=1-p=\frac{12}{13} \)
\( \displaystyle P(X=0)=q^2=\frac{144}{169} \)
\( \displaystyle P(X=1)=2pq \)
\( \displaystyle =2\cdot\frac{1}{13}\cdot\frac{12}{13} \)
\( \displaystyle =\frac{24}{169} \)
\( \displaystyle P(X=2)=p^2=\frac{1}{169} \)
Check: \( \frac{144+24+1}{169}=1 \).
Why the other options are incorrect
Option (2) \( \begin{pmatrix}0 & 1 & 2\\\displaystyle\frac{1}{169} & \displaystyle\frac{24}{169} & \displaystyle\frac{144}{169}\end{pmatrix} \) — Incorrect. The probabilities of X = 0 and X = 2 are interchanged; getting no queen is far more likely than getting two queens.
Option (3) \( \begin{pmatrix}0 & 1 & 2\\\displaystyle\frac{144}{169} & \displaystyle\frac{1}{169} & \displaystyle\frac{24}{169}\end{pmatrix} \) — Incorrect. The probabilities of X = 1 and X = 2 are interchanged; one queen can come in two orders, so it is 24 times as likely as two queens.
Option (4) \( \begin{pmatrix}0 & 1 & 2\\\displaystyle\frac{144}{169} & \displaystyle\frac{2}{169} & \displaystyle\frac{23}{169}\end{pmatrix} \) — Incorrect. \( P(X=1) \) is \( 2pq=\frac{24}{169} \), not \( \frac{2}{169} \), and \( P(X=2) \) is \( \frac{1}{169} \), not \( \frac{23}{169} \).