Question

The face cards are removed from a full pack. Out of the remaining 40cards, 4 are drawn at random. What is the probability that they belong to different suits?
A. $\dfrac{{{\text{1000}}}}{{{\text{9139}}}}$
B. $\dfrac{{{\text{1001}}}}{{{\text{9136}}}}$
C. $\dfrac{{{\text{100}}}}{{{\text{913}}}}$
D. $\dfrac{{{\text{1002}}}}{{{\text{9129}}}}$

Answer 1

We use concept of combinations to solve this problem. After removing the face cards, each suite has 10 cards. The probability of getting 4 cards of each suit is given by the possible number of ways of selecting one card of each suit divided by the total number of ways of selecting 4 cards from 40 cards. Number of ways of selecting r cards from n cards is given by ${}^{\text{n}}{{\text{C}}_{\text{r}}} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$

Complete step by step Answer:

We need to select 4 cards from a deck of 40 cards, which is after the face cards are removed.
We know that number of ways of selecting r objects from n objects is given by ${}^{\text{n}}{{\text{C}}_{\text{r}}} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Total number of ways of selecting 4 cards from 40 cards is given by, ${}^{{\text{40}}}{{\text{C}}_{\text{4}}}$.
On calculating the combinations, we get,
${}^{{\text{40}}}{{\text{C}}_{\text{4}}}{\text{ = }}\dfrac{{{\text{40!}}}}{{{\text{4!}}\left( {{\text{36}}} \right){\text{!}}}}$
By simplification, we get
{\text{ = }}\dfrac{{{\text{40 \times$39$\times$38$\times37}}}}{{{\text{4\times$3$\times 2}}}}{\text{ = 91390}}
Therefore, total number of ways of selecting 4 cards from 40 cards is 91390.
Now we need to find the number of ways of selecting 1 card from each suit of 10 cards, and we have 4 such suits.
Therefore, number of ways of selecting 4 cards of each suit is given by,
${}^{{\text{10}}}{{\text{C}}_{\text{1}}}{\text{ $\times$ }}{}^{{\text{10}}}{{\text{C}}_{\text{1}}}{\text{ $\times$ }}{}^{{\text{10}}}{{\text{C}}_{\text{1}}}{\text{ $\times$ }}{}^{{\text{10}}}{{\text{C}}_{\text{1}}}{\text{ = }}{\left( {{}^{{\text{10}}}{{\text{C}}_{\text{1}}}} \right)^{\text{4}}}$
Using combinations, we get,
${\text{ = }}{\left( {\dfrac{{{\text{10!}}}}{{{\text{1!}}\left( {\text{9}} \right){\text{!}}}}} \right)^{\text{4}}}{\text{ = 1}}{{\text{0}}^{\text{4}}}{\text{ = 10000}}$
The number of ways of selecting 4 cards of each suit is 10000
The required probability is given by $\dfrac{{{\text{The number of ways of selecting 4 cards of each suit}}}}{{{\text{, total number of ways of selecting 4 cards from 40 cards}}}}$
${\text{ = }}\dfrac{{{\text{10000}}}}{{{\text{91390}}}}{\text{ = }}\dfrac{{{\text{1000}}}}{{{\text{9139}}}}$
Therefore, the required probability is $\dfrac{{{\text{1000}}}}{{{\text{9139}}}}$
So, the correct answer is option A.

Note: The concept of permutations and combinations are used here to find the probability. We must be aware of the different suits, types of cards, and their numbers. We must not use permutations to find the number of ways of selecting. We use permutations if the order is important. We must take the product of the combinations of selecting 1 card from each suit to get the total combination and not the sum. The probability of an event is defined as the number of favourable outcomes divided by the total number of outcomes.