Question

Two cards are drawn at random from a pack of 52 cards. Find the probability that
a) One is a face card and the other is an ace card.
b) One is a club and the other is diamond.
c) Both are from the same suit.
d) Both are red cards.
e) One is a heart card and the other is a non-heart card.

Answer 1

To solve this question, we should know the composition of a pack of 52 cards. We know that there are four suits in a pack of cards and each suit has 13 cards in them. Out of those 13 cards, there will be 3 face cards per suit and one ace card and the remaining 9 cards number from 2 to 10. Out of the four suits, hearts and diamonds are red in colour and spades and clubs are black in colour. To select an item out of n items, we can write the number of ways as ${}^{n}{{C}_{1}}=n$. We can write the number of ways to select r items out of n items as ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. Using these two formulae and the composition of the cards, we can get the required probabilities of different events.

Complete step-by-step answer:
We are given a pack of standard 52 cards. We know that there are four suits in a pack of cards and each suit has 13 cards in them. Out of those 13 cards, there will be 3 face cards per suit and one ace card and the remaining 9 cards number from 2 to 10. Out of the four suits, hearts and diamonds are red in colour and spades and clubs are black in colour.
We can write the number of ways to select r items out of n items as ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.
We are given that two cards are drawn at random from the pack of 52 cards. We can write the total number of ways from the above formula as
$\begin{aligned} & n\left( S \right)={}^{52}{{C}_{2}}=\dfrac{52!}{2!\left( 52-2 \right)!}=\dfrac{52!}{2!\times 50!}=\dfrac{52\times 51}{2}=26\times 51 \\\ & n\left( S \right)=26\times 51 \\\ \end{aligned}$
Let us consider the first event.
Let us consider the event ${{E}_{1}}$ as one is a face card and the other is an ace card. There are 3 face cards per suit and 4 suits. So, the total number of face cards is 12. There is an ace card per suit. So, there are 4 ace cards in total. So, we can write that the number of ways the event ${{E}_{1}}$ can happen is the product of selecting 1 card from 12 face cards and 1 card from 4 ace cards. We can write it as
$n\left( {{E}_{1}} \right)={}^{12}{{C}_{1}}\times {}^{4}{{C}_{1}}=12\times 4=48$
We know that $P\left( {{E}_{1}} \right)=\dfrac{n\left( {{E}_{1}} \right)}{n\left( S \right)}$. We can write that
$P\left( {{E}_{1}} \right)=\dfrac{48}{26\times 51}=\dfrac{24}{13\times 51}=\dfrac{8}{13\times 17}=\dfrac{8}{221}$
Let us consider the event ${{E}_{2}}$ as one is a club and the other is a diamond. We know that the total number of club cards is 13. We know that the total number of diamond cards is 13. So, we can write that the number of ways the event ${{E}_{2}}$ can happen is the product of selecting 1 card from 13 club cards and 1 card from 13 diamond cards. We can write it as
$n\left( {{E}_{2}} \right)={}^{13}{{C}_{1}}\times {}^{13}{{C}_{1}}=13\times 13=169$
We can write that
$P\left( {{E}_{2}} \right)=\dfrac{169}{26\times 51}=\dfrac{13}{2\times 51}=\dfrac{13}{102}$
Let us consider the event ${{E}_{3}}$ be when both the cards are from the same suit. We know that there are 13 cards in a suit and there are 4 suits in total. So, we can write that the number of ways the event ${{E}_{3}}$ can happen is the product of selecting 1 suit from 4 suits and 2 cards from 13 cards of that suit.
$n\left( {{E}_{3}} \right)={}^{4}{{C}_{1}}\times {}^{13}{{C}_{2}}=4\times \dfrac{13!}{2!\left( 13-2 \right)!}=4\times \dfrac{13\times 12}{2}=13\times 24$
$P\left( {{E}_{3}} \right)=\dfrac{13\times 24}{26\times 51}=\dfrac{24}{2\times 51}=\dfrac{12}{51}$
Let us consider the event ${{E}_{4}}$be when both the cards are red. There are 26 red cards in total. So, we can write that the number of ways the event ${{E}_{4}}$ can happen is by selecting 2 cards from 26 red cards.
$n\left( {{E}_{4}} \right)={}^{26}{{C}_{2}}=\dfrac{26!}{2!\left( 26-2 \right)!}=\dfrac{26\times 25}{2}=13\times 25$
$P\left( {{E}_{4}} \right)=\dfrac{13\times 25}{26\times 51}=\dfrac{25}{2\times 51}=\dfrac{25}{102}$
Let us consider the event ${{E}_{5}}$ as one is a heart card and the other not a heart card. There are 13 heart cards and 39 other than heart cards. So, we can write that the number of ways the event ${{E}_{5}}$ can happen is the product of selecting 1 card from 13 heart cards and 1 card from 39 other than heart cards. We can write it as
$n\left( {{E}_{5}} \right)={}^{13}{{C}_{1}}\times {}^{39}{{C}_{1}}=13\times 39$
$P\left( {{E}_{5}} \right)=\dfrac{13\times 39}{26\times 51}=\dfrac{39}{2\times 51}=\dfrac{39}{102}$
$\therefore$Hence, we found out the probabilities for different cases.

Note: We should observe that the order of selection is not considered in this question. If the order of selection is considered, the number of cases for every event will be multiplied by 2. In the event ${{E}_{3}}$, some students forget to multiply ${}^{13}{{C}_{2}}$ by 4 in a hurry. We should understand exactly what is asked. We need to get two cards of the same suit but the suit can be anyone. So, we should multiply it by 4. We should understand the question properly to not make a mistake in these types of questions.