Question

Two cards are drawn at random from a pack of 52 cards. What is the probability that it will be
a). A club and a heart
b). A king and a queen
c).two kings
d).two cards from the same suit.

Answer 1

Hint: From the pack of 52 cards 2 were taken out randomly. Therefore, the total number of outcomes will be ;
$$\text{total number of outcome}{=}^{52}{C}_2={\displaystyle \frac{52!}{2!(52-2)!}}={\displaystyle \frac{52!}{(2!)(50!)}}={\displaystyle \frac{52\times 51}{2}}=1326$$ which will not change throughout the sums.
We will make favorable outcomes according to the question given one by one and solve the question accordingly. Suppose, for a club and a heart we know that these to two are two different suits therefore we will choose them separately $${}^{13}{C}_1{\times }^{13}{C}_1$$.
Formula used:
$$\text{probability}={\displaystyle \frac{\text{total favourable outcome}}{\text{total number of outcome}}}$$
$${}^n{C}_r={\displaystyle \frac{n!}{r!(n-r)!}}$$.


Complete step-by-step answer:
a). There are 13 cards in the club and 13 cards in the heart. We need to pick two cards at random so we will pick one card from each suit.

$$\text{favourable outcome}{=}^{13}{C}_1{\times }^{13}{C}_1\text{probability}={\displaystyle \frac{{}^{13}{C}_1{\times }^{13}{C}_1}{{}^{52}{C}_2}}={\displaystyle \frac{13\times 13}{{\displaystyle \frac{52!}{2!(52-2)!}}}}={\displaystyle \frac{13\times 13}{1326}}\text{=0}\text{.12745 }{[}^n{\text{C}}_1\text{=n }]$$
Therefore, the probability that it will be a club and a heart will be 0.12745.

b). There are 4 kings in 52 cards. Therefore, the favourable number of outcome will be
$$\text{favourable outcome}{=}^4{C}_2$$
probability of getting two kings are :
$$\text{probability}={\displaystyle \frac{{}^4{C}_2}{{}^{52}{C}_2}}={\displaystyle \frac{6}{1326}}=0.004524[\text{ using}{}^n{C}_r={\displaystyle \frac{n!}{r!(n-r)!}}]$$

c). There are 4 kings and 4 queens but we need only 1 king and 1 queen. Therefore, the favourable number of outcomes will be $$\text{favourable outcome}{=}^4{C}_1{\times }^4{C}_1$$
probability....
$$\text{probability}={\displaystyle \frac{{}^4{C}_1{\times }^4{C}_1}{{}^{52}{C}_2}}={\displaystyle \frac{4\times 4}{1326}}=0.012066{[}^n{\text{C}}_1\text{=n};\text{usnig }{\displaystyle \frac{n!}{r!(n-r)!}}]$$

 Therefore, the probability that it will be a king and a king will be 0.01206

d). Here, we first need to choose a suit. Since we don’t know the suit we need to pick out a suit from the four suits and from that suit we will pick 2 cards from the 13 cards.
Therefore the favorable number of outcomes is...
$$\text{favourable outcome}{=}^4{C}_1{\times }^{13}{C}_2[\text{ selecting the suit then picking 2 cards }]$$

Probability is.....
$$\text{probability}={\displaystyle \frac{{}^4{C}_1{\times }^{13}{C}_2}{{}^{52}{C}_2}}={\displaystyle \frac{4\times 78}{1326}}=0.2352{[}^n{\text{C}}_1\text{=n};\text{usnig }{\displaystyle \frac{n!}{r!(n-r)!}}]$$

Therefore, the probability that it will be 2 cards from the same suit will be 0.2352.

Note: In question subpart b) and c) we do not choose a suit first because there are only 4 kings and queens. Picking out any suit and then choosing the king or queen will yield us the wrong answer since all the king and queen are the same unlike the normal cards, including an ace, where they have different colors and shapes of their suit.
In subpart d) we will need to pick out a separate suit first because we ‘maybe’ get normal cards or king or queen. Therefore, it is wise to choose a suit and then select the cards.