Q78 / 150 · MHT CET 2023 (12 May Shift 1) · Mathematics

Two cards are drawn successively with replacement from a well shuffled pack of 52 cards. Then the probability distribution of number of jacks is

  • Correct answer
    A
    \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}
  • B
    \begin{array}{|l|c|c|c|} \hline \mathrm{X}=x & 0 & 1 & 2 \\ \hline \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 \\ \hline \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 \\ \hline \mathrm{P}(\mathrm{X}=x) & \frac{144}{169} & \frac{1}{169} & \frac{24}{169} \\ \hline \end{array}

Solution

Let $\mathrm{X}$ denotes the number of jacks $\therefore \quad$ Possible values of $\mathrm{X}$ are $0,1,2$ $$ \therefore \quad \mathrm{P}(\mathrm{X}=0)=\frac{{ }^{48} \mathrm{C}_1 \times{ }^{48} \mathrm{C}_1}{{ }^{52} \mathrm{C}_1 \times{ }^{52} \mathrm{C}_1}=\frac{144}{169} $$ $$ \begin{aligned} & \mathrm{P}(\mathrm{X}=1)=\frac{{ }^{48} \mathrm{C}_1 \times{ }^4 \mathrm{C}_1}{{ }^{52} \mathrm{C}_1 \times{ }^{52} \mathrm{C}_1}+\frac{{ }^4 \mathrm{C}_1 \times{ }^{52} \mathrm{C}_1}{{ }^{52} \mathrm{C}_1 \times{ }^{52} \mathrm{C}_1}=\frac{24}{169} \\ & \mathrm{P}(\mathrm{X}=2)=\frac{{ }^4 \mathrm{C}_1 \times{ }^4 \mathrm{C}_1}{{ }^{52} \mathrm{C}_1 \times{ }^{52} \mathrm{C}_1}=\frac{1}{169} \end{aligned} $$ $\therefore \quad$ Option (A) is correct.

Hence, the correct answer is option: 'A'.

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