Question

In a card game using a standard \[52\] card deck, \[4\]-card hands are dealt. What’s the probability of being dealt \[3\] diamonds ?

Answer 1

Hint:To find the probability of being dealt \[3\] diamonds, we will use the concept of combination. We will first find the number of ways selecting \[3\] diamonds cards from \[13\] diamonds cards and then the number of ways in which fourth card will not be a diamond in order to find total ways in which \[4\]-card hands are dealt such that being dealt \[3\] diamonds. We will then find the total number of ways in which \[4\]-card hands can be dealt from \[52\] card deck. To find the probability of being dealt \[3\] diamonds, we will take ratio of total ways in which \[4\]-card hands are dealt such that being dealt \[3\] diamonds to total number of ways in which \[4\]-card hands can be dealt from \[52\] card deck.

Complete step by step answer:
We have to find the number of possible \[4\]-card hands which contain exactly \[3\] diamonds.As we know, in a standard deck there are \[13\] ordinal cards i.e., Jack, Queen, King and \[10\] aces and in each of \[4\] suits i.e., Hearts, Diamond, Clubs and Spades.Therefore, we have total \[13 \times 4\] cards i.e., \[52\] cards.We will use the concept of Combination to determine the number of possible \[4\]-card hands which contain exactly \[3\] diamonds as the order in which we are selecting the cards is not important here. So, we can use the concept of combination to solve this problem.As we know,
\[^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]
where \[n\] is the number of items and \[r\] is the number of items being chosen at a time.

Number of ways in which \[3\] diamonds can be selected from \[13\] diamonds card will be \[^{13}{C_3}\] i.e.,
Number of ways of selecting \[3\] diamonds cards from \[13\] diamonds cards \[ = \] \[^{13}{C_3}\]
Putting in formula we get
\[{ \Rightarrow ^{13}}{C_3} = \dfrac{{13!}}{{3!\left( {13 - 3} \right)!}}\]
On expanding we get
\[{ \Rightarrow ^{13}}{C_3} = \dfrac{{13 \times 12 \times 11 \times 10!}}{{3 \times 2 \times 10!}}\]

On calculation we get
\[{ \Rightarrow ^{13}}{C_3} = 286\]
So, the number of ways of selecting \[3\] diamonds cards from \[13\] diamonds cards \[ = \]\[286\]
The fourth card can’t be a diamond. Therefore, we have to select from the remaining suits.
Number of ways in which fourth card will not be a diamond \[ = \] \[52 - 13\]
\[ = 39\]
Putting all together we get, number of ways of being dealt \[3\] diamonds \[ = \] \[286 \times 39\]\[ = 11154\]
Number of ways in which \[4\] card hands can be dealt from \[52\] cards \[{ = ^{52}}{C_4}\]\[ = \dfrac{{52!}}{{\left( {52 - 4} \right)!4!}}\]

On expanding, we get
Number of ways in which \[4\] card hands can be dealt from \[52\] cards \[ = \dfrac{{52 \times 51 \times 50 \times 49 \times 48!}}{{48! \times 4 \times 3 \times 2 \times 1}}\]
On simplification, we get
Number of ways in which \[4\] card hands can be dealt from \[52\] cards \[ = 270125\]
The probability of being dealt \[3\] diamonds \[ = \dfrac{{{\text{number of ways of being dealt 3 diamonds}}}}{{{\text{Total number of ways in which 4 card hands can be dealt from 52 cards}}}}= \dfrac{{11154}}{{270125}}\]
On simplification we get
The probability of being dealt \[3\] diamonds \[ = 0.041\]

Therefore, the probability of being dealt \[3\] diamonds is \[0.041\].

Note:Here, we are not interested in the order of the draw, that’s why we have used combination because a combination is a grouping or subset of items where the order does not matter. But if the order did matter then it will be solved using permutation as permutation is an arrangement in a definite order.