An urn contains 5 white and …

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}