Question

In bridge game of playing cards, $4$ players are distributed one card each by turn so that each player gets $13$ cards. Find out the probability of a specified player getting a black ace and a king.

• A. $p=\frac{82251}{978775}$
• B. $p=\frac{164502}{978775}$
• C. $p=\frac{329004}{978775}$
• D. $p=\frac{82251}{1957550}$

Answer 1

Answer: (B), (C)

$p=\frac{164502}{978775}$
$p=\frac{329004}{978775}$

Since, each player gets 13 cards.
So, n=total no. of ways =${}^{52}{C}_{13}=\frac{52!}{13!39!}$
Now, there are 2 black aces and 4 kings in a deck of 52 cards. Since one black ace out of 2 black aces can be chosen in ${}^2{C}_1$ways, one king out of 4 kings in ${}^4{C}_1$ ways and the remaining 11 cards from the remaining 46 cards in ${}^{46}{C}_{11}$ ways.
So, $m={}^2{C}_1\times {}^4{C}_1\times {}^{46}{C}_{11}$,
Required probability $p=\frac{m}{n}=\frac{{}^2{C}_1{\times }^4{C}_1{\times }^{46}{C}_{11}}{{}^{52C_{13}}}$