Two cards are drawn successively with replacement from well shuffled pack of 52 cards, then the probability…
Two cards are drawn successively with replacement from well shuffled pack of 52 cards, then the probability distribution of number of queens is
- $\begin{array}{|l|c|c|c|}\hline \mathrm{X}=x & 0 & 1 & 2 \\\hline \mathrm{P}[\mathrm{X}=x] & \frac{144}{169} & \frac{24}{169} & \frac{1}{169} \\\hline\end{array}$
- $\begin{array}{|l|c|c|c|}\hline X=x & 0 & 1 & 2 \\\hline P[X=x] & \frac{1}{169} & \frac{24}{169} & \frac{144}{169} \\\hline\end{array}$
- $\begin{array}{|l|c|c|c|}\hline \mathrm{X}=x & 0 & 1 & 2 \\\hline \mathrm{P}[\mathrm{X}=x] & \frac{24}{169} & \frac{1}{169} & \frac{144}{169} \\\hline\end{array}$
- $\begin{array}{|l|c|c|c|}\hline \mathrm{X}=x & 0 & 1 & 2 \\\hline \mathrm{P}[\mathrm{X}=x] & \frac{1}{169} & \frac{25}{169} & \frac{143}{169} \\\hline\end{array}$
Solution
Let $\mathrm{X}$ denote the number of queens.
$\therefore \quad$ Possible values of $\mathrm{X}$ are $0,1,2$.
$\begin{aligned}
& \mathrm{P}(\text { queen })=\frac{4}{52}=\frac{1}{13} \\
& \begin{aligned}
\mathrm{P}(\text { not a queen }) & =\frac{48}{52}=\frac{12}{13} \\
\mathrm{P}(\mathrm{X}=0) & =\frac{12}{13} \times \frac{12}{13} \\
& =\frac{144}{169} \\
\mathrm{P}(\mathrm{X}=1) & =\left(\frac{1}{13} \times \frac{12}{13}\right)+\left(\frac{12}{13} \times \frac{1}{13}\right) \\
& =\frac{12}{169}+\frac{12}{169}=\frac{24}{169} \\
\mathrm{P}(\mathrm{X}=2) & =\frac{1}{13} \times \frac{1}{13} \\
& =\frac{1}{169}
\end{aligned}
\end{aligned}
$
$\begin{array}{|c|c|c|c|}
\hline \mathrm{X}=x & 0 & 1 & 2 \\
\hline \mathrm{P}[\mathrm{X}=x] & \frac{144}{169} & \frac{24}{169} & \frac{1}{169} \\
\hline
\end{array}$
The probability distribution of $\mathrm{X}$ is
Asked in: MHT CET 2023 (14 May Shift 1)
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