Question

In a 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) \[p=\dfrac{82251}{978775}\]
b) \[p=\dfrac{164502}{978775}\]
c) \[p=\dfrac{329004}{978775}\]
d) \[p=\dfrac{82251}{1957550}\]

Answer 1

Hint:The basic concept of probability states that probability of an event is equal to the ratio of total number of favourable outcomes to the total number of the outcome, that is, \[\text{probability of an event}=\dfrac{\text{total number of favourable outcome}}{\text{total number of outcomes}}\].

Complete step-by-step answer:
In this question, we have to find the probability of a specified player a card of black ace and a king.
As given in the question, we can say, the total number of cards each player will get are 13. So, the number of ways in which a player can get 13 cards are equal to \[{}^{52}{{C}_{13}}\] where 52 represent the total number of cards and 13 represents the total number of cards to be selected.
Therefore, we can say the total number of outcomes of distributing 13 cards to a specified player is \[{}^{52}{{C}_{13}}=\dfrac{52!}{13!39!}\].
Now, we know that a deck of cards contains 2 black aces and 4 kings, so the number of ways of choosing one out of these 2 black aces is \[{}^{2}{{C}_{1}}\] and the number of ways of choosing one out of four kings is \[{}^{4}{{C}_{1}}\].
Now, we can say that the total number of ways of giving rest of the 11 cards out of remaining 46 cards (do not include any of the kings and black aces) is \[{}^{46}{{C}_{11}}\].
Now, as we know that \[\text{probability of an event}=\dfrac{\text{total number of favourable outcome}}{\text{total number of outcomes}}\]
So, let the probability of distributing 13 cards to a specified player including one black ace and one king be ‘p’ is equal to,
\[p=\dfrac{{}^{2}{{C}_{1}}\times {}^{4}{{C}_{1}}\times {}^{46}{{C}_{11}}}{{}^{52}{{C}_{13}}}\]
For further simplifying, we will use, \[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
Therefore, we will get, \[p=\dfrac{2\times 4\times \dfrac{46!}{11!\left( 46-11 \right)!}}{\dfrac{52!}{13!\left( 52-13 \right)!}}\]
On further simplification, we will get,
\[p=\dfrac{2\times 4\times \dfrac{46!}{11!35!}}{\dfrac{52!}{13!39!}}\]
\[p=\dfrac{2\times 4\times 13!\times 39!\times 46!}{52!\times 11!\times 35!}\]
\[p=\dfrac{2\times 4\times 13\times 12\times 39\times 38\times 37\times 36}{52\times 51\times 50\times 49\times 48\times 47}\]
\[p=\dfrac{54834}{70825}\]
Hence, the probability of distributing 13 cards to a specified player including one black ace and one king is \[p=\dfrac{54834}{70825}\].

Note: In this question, we have frequently used the formula of \[{}^{n}{{C}_{r}}\], which is equal to \[\dfrac{n!}{r!\left( n-r \right)!}\], because when we have to choose few out of all in a particular manner we apply formula of combination, which gives us the maximum possible cases of choosing them. In this question, we have to find the possible case of giving cards to players, so we used it.