Question
Two cards are drawn from a pack of well shuffled cards. Find the probability that one is a club and other in King.
Answer: Option D
Let X be the event that cards are in a club which is not king and other is the king of club.
Let Y be the event that one is any club card and other is a non-club king.
Hence, required probability:
$$\eqalign{
& = P(A) + P(B) \cr
& = \frac{{^{12}{C_1}{ \times ^1}{C_1}}}{{^{52}{C_2}}} + \frac{{^{13}{C_1}{ \times ^3}{C_1}}}{{^{52}{C_2}}} \cr
& = \left( {\frac{{2 \times \left( {12 \times 1} \right)}}{{52 \times 51}}} \right) + \left( {\frac{{2\left( {13 \times 3} \right)}}{{52 \times 51}}} \right) \cr
& = \left( {\frac{{24 + 78}}{{52 \times 51}}} \right) \cr
& = \frac{1}{{26}} \cr} $$
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Let X be the event that cards are in a club which is not king and other is the king of club.
Let Y be the event that one is any club card and other is a non-club king.
Hence, required probability:
$$\eqalign{
& = P(A) + P(B) \cr
& = \frac{{^{12}{C_1}{ \times ^1}{C_1}}}{{^{52}{C_2}}} + \frac{{^{13}{C_1}{ \times ^3}{C_1}}}{{^{52}{C_2}}} \cr
& = \left( {\frac{{2 \times \left( {12 \times 1} \right)}}{{52 \times 51}}} \right) + \left( {\frac{{2\left( {13 \times 3} \right)}}{{52 \times 51}}} \right) \cr
& = \left( {\frac{{24 + 78}}{{52 \times 51}}} \right) \cr
& = \frac{1}{{26}} \cr} $$
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