Question

: From a well shuffled pack of \[52\] cards, one card is drawn at random. Find the probability of getting a diamond.

Answer 1

Hint: There are total \[52\] cards in a deck with \[4\] different suits: ace, diamonds, hearts and clubs. All the suits have equal number of cards and each suit has \[13\] cards. Hence to arrive at the probability, we will have to divide \[13\] by \[52\] with the formula \[P(E) = \dfrac{{Number\,of\,favourable\,outcomes}}{{Total\,number\,of\,outcomes}}\] .

Complete step-by-step answer:

We are given the figure as follows, represented as a well shuffled pack of \[52\] cards, one card is drawn at random. Find the probability of getting a diamond.
We will use combination techniques to find out the solution. A combination is a mathematical technique for calculating the number of possible arrangements in a collection of items where the order of the items is irrelevant. It is denoted as \[n{C_r}\] where \[n\] is the total number of objects and \[r\] is the number of selections.
The formula to find combination is given as follows:
 \[n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}\]
Moreover, probability of an event can be obtained as ratio of occurrence of an event to the total sample size. The formula of finding probability of a favourable event is \[P(E) = \dfrac{{Number\,of\,favourable\,outcomes}}{{Total\,number\,of\,outcomes}}\]
Now we can proceed to solve the sum as follows:
Total cards in pack are \[52\] . Hence the total combination of selecting one card will be-
 \[n{C_r} = \dfrac{{52!}}{{1!(52 - 1)!}}\]
 \[ = \dfrac{{52!}}{{1!(51)!}}\]
 \[ = \dfrac{{52 \times 51!}}{{1!(51)!}}\]
 \[ = 51{C_1}\]
Total number of diamonds in a pack are \[13\] starting from A to \[10\] , King, Queen and Jack. Hence the total combination of selecting one card of diamond will be-
 \[n{C_r} = \dfrac{{13!}}{{1!(13 - 1)!}}\]
 \[ = \dfrac{{13!}}{{1!(12)!}}\]
 \[ = \dfrac{{13 \times 12!}}{{1!(12)!}}\]
 \[ = 13{C_1}\]
Now let the probability of getting a diamond be \[P(D)\] :
 \[P(D) = \dfrac{{Number\,of\,getting\,diamond}}{{Total\,number\,of\,cards}}\]
 \[ = \dfrac{{13{C_1}}}{{52{C_1}}}\]
 \[ = \dfrac{{13}}{{52}}\]
 \[ = \dfrac{1}{4}\]
In terms of percentage, it will be \[\dfrac{1}{4} \times 100 = 25\% \]
So, the correct answer is “ \[ = \dfrac{1}{4}\] ”.

Note: We can solve the sum directly without applying the combination formula since there is selection of only one card.
We know that there are \[13\] diamond cards so favourable outcomes will be \[13\] and sample size will be \[52\] as the total number of cards. We use the probability formula to obtain the required solution.