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Preeti Dabral 3 years, 3 months ago
Step 1 of 2
Two cards are drawn simultaneously,
Let, denotes the number of aces in 2 draws from a pack of 52 cards
∴ X can takes values 0, 1, 2
Step 2
{tex}\begin{aligned} & P(X=0)=\frac{{ }^{48} C_2}{{ }^{52} C_2}=\frac{48 \times 47}{52 \times 51}=\frac{12 \times 47}{13 \times 51}=\frac{188}{221} \\ & \therefore P(X=0)=\frac{188}{221} \\ & P(X=1)=\frac{{ }^4 C_1 \times{ }^{48} C_1}{{ }^{52} C_2}=\frac{4 \times 48 \times 2}{52 \times 51}=\frac{32}{221} \\ & \therefore P(X=1)=\frac{32}{221} \\ & P(X=2)=\frac{{ }^4 C_2}{{ }^{52} C_2}=\frac{4 \times 3}{52 \times 51}=\frac{1}{221} \\ & \therefore P(X=2)=\frac{1}{221} \\ & \end{aligned}{/tex}
Thus the probability distribution of random variable is
{tex}\begin{array}{|l|l|l|l|}
\hline X & 0 & 1 & 2 \\
\hline P(X) & \frac{188}{221}= & \frac{32}{221}= & \frac{1}{221}= \\
\hline
\end{array}{/tex}
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Mohammad Shish 4 years, 7 months ago
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