MHT CET · Maths · Probability
Two cards are drawn successively with replacement from a well shuffled pack of 52 cards. Then the probability distribution of number of jacks is
- A \(\begin{array}{|l|c|c|c|}
\hline \mathrm{X}=x & 0 & 1 & 2\mathrm{P}(\mathrm{X}=x) & \frac{144}{169} & \frac{24}{169} & \frac{1}{169} \\
\hline
\end{array}\) - B \(\begin{array}{|l|c|c|c|}
\hline \mathrm{X}=x & 0 & 1 & 2\mathrm{P}(\mathrm{X}=x) & \frac{1}{169} & \frac{144}{169} & \frac{24}{169} \\
\hline
\end{array}\) - C \(\begin{array}{|l|c|c|c|}
\hline \mathrm{X}=x & 0 & 1 & 2\mathrm{P}(\mathrm{X}=x) & \frac{24}{169} & \frac{1}{169} & \frac{144}{169} \\
\hline
\end{array}\) - D \(\begin{array}{|c|c|c|c|}
\hline \mathrm{X}=x & 0 & 1 & 2\mathrm{P}(\mathrm{X}=x) & \frac{144}{169} & \frac{1}{169} & \frac{24}{169}\\ \hline\end{array}\)
Answer & Solution
Correct Answer
(A) \(\begin{array}{|l|c|c|c|}
\hline \mathrm{X}=x & 0 & 1 & 2\mathrm{P}(\mathrm{X}=x) & \frac{144}{169} & \frac{24}{169} & \frac{1}{169} \\
\hline
\end{array}\)
Step-by-step Solution
Detailed explanation
Let \(\mathrm{X}\) denotes the number of jacks
\(\therefore\) Possible values of \(\mathrm{X}\) are \(0,1,2\)
\(\therefore \mathrm{P}(\mathrm{X}=0)=\frac{{ }^{48} \mathrm{C}_1 \times{ }^{48} \mathrm{C}_1}{{ }^{52} \mathrm{C}_1 \times{ }^{52} \mathrm{C}_1}=\frac{144}{169}\)
\(\mathrm{P}(\mathrm{X}=1)=\frac{{ }^{48} \mathrm{C}_1 \times{ }^4 \mathrm{C}_1}{{ }^{52} \mathrm{C}_1 \times{ }^{52} \mathrm{C}_1}+\frac{{ }^4 \mathrm{C}_1 \times{ }^{52} \mathrm{C}_1}{{ }^{52} \mathrm{C}_1 \times{ }^{52} \mathrm{C}_1}=\frac{24}{169} \)
\( \mathrm{P}(\mathrm{X}=2)=\frac{{ }^4 \mathrm{C}_1 \times{ }^4 \mathrm{C}_1}{{ }^{52} \mathrm{C}_1 \times{ }^{52} \mathrm{C}_1}=\frac{1}{169}\)
\(\therefore\) Option (A) is correct.
\(\therefore\) Possible values of \(\mathrm{X}\) are \(0,1,2\)
\(\therefore \mathrm{P}(\mathrm{X}=0)=\frac{{ }^{48} \mathrm{C}_1 \times{ }^{48} \mathrm{C}_1}{{ }^{52} \mathrm{C}_1 \times{ }^{52} \mathrm{C}_1}=\frac{144}{169}\)
\(\mathrm{P}(\mathrm{X}=1)=\frac{{ }^{48} \mathrm{C}_1 \times{ }^4 \mathrm{C}_1}{{ }^{52} \mathrm{C}_1 \times{ }^{52} \mathrm{C}_1}+\frac{{ }^4 \mathrm{C}_1 \times{ }^{52} \mathrm{C}_1}{{ }^{52} \mathrm{C}_1 \times{ }^{52} \mathrm{C}_1}=\frac{24}{169} \)
\( \mathrm{P}(\mathrm{X}=2)=\frac{{ }^4 \mathrm{C}_1 \times{ }^4 \mathrm{C}_1}{{ }^{52} \mathrm{C}_1 \times{ }^{52} \mathrm{C}_1}=\frac{1}{169}\)
\(\therefore\) Option (A) is correct.
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