question_answer

Two cards are drawn successively without   replacement   from a Well-shuffled pack of 52 cards. Find the probability distribution of ace cards.

Answer:

\[\because \] Total number of cards in a pack = 52 where, ace cards = 4, other cards = 48 Let X be a random variable that denotes the number of ace cards in a successive drawn of two cards. Then, X can take values 0, 1 and 2. Now, P(X = 0) = P (no ace card is drawn) \[=\frac{^{4}{{C}_{0}}{{\times }^{48}}{{C}_{2}}}{^{52}{{C}_{2}}}=\frac{1\times \frac{48\times 47}{2\times 1}}{\frac{52\times 51}{2\times 1}}=\frac{48\times 47}{52\times 51}=\frac{188}{221}\] P(X = 1) = P (one ace card is drawn) \[=\frac{^{4}{{C}_{1}}{{\times }^{48}}{{C}_{1}}}{^{52}{{C}_{2}}}=\frac{4\times 48}{\frac{52\times 51}{2\times 1}}\] \[=\frac{4\times 48\times 2\times 1}{52\times 51}=\frac{32}{221}\]           P(X = 2)= P(both ace cards are drawn) \[=\frac{^{4}{{C}_{2}}{{\times }^{48}}{{C}_{0}}}{^{52}{{C}_{2}}}=\frac{\frac{4\times 3}{2\times 1}\times 1}{\frac{52\times 51}{2\times 1}}=\frac{4\times 3}{52\times 51}=\frac{1}{221}\] \[\therefore \] Required probability distribution is                 

X	0	1	2
P(X)	\[\frac{188}{221}\]	\[\frac{32}{221}\]	\[\frac{1}{221}\]