Question

Find the probability of getting different suit cards and different denomination cards when two cards are drawn from the pack-
A.$\dfrac{{13}}{{17}}$
B.$\dfrac{{13}}{{34}}$
C.$\dfrac{{12}}{{17}}$
D.$\dfrac{6}{{17}}$

Answer 1

Hint: Here we have to choose two cards from the pack of $52$ cards which contain $4$ suits (heart, club, spade, and diamond) with $13$ cards each. So we can use the following formula to select the two cards-The number of ways to select r things from n things=${}^n{C_r}$ where n= total number of things and r is the number of things to be selected. And also we know that the formula of combination is-
${}^n{C_r} = \dfrac{{n!}}{{r!n - r!}}$.Now use the formula of probability which is given as-
$ \Rightarrow $ P=$\dfrac{{{\text{required ways to select 2 cards}}}}{{{\text{total ways to select cards}}}}$

Complete step by step answer:

Given, two cards are drawn from the pack. We know that the pack has $52$ cards then the number of ways the $2$ cards can be selected from $52$ cards =${}^{52}{C_2}$
Then total ways to select two cards=${}^{52}{C_2}$
We know that ${}^n{C_r} = \dfrac{{n!}}{{r!n - r!}}$
Then we get
$ \Rightarrow $ ${}^{52}{C_2} = \dfrac{{52!}}{{2!52 - 2!}}$
On solving we get,
$ \Rightarrow {}^{52}{C_2} = \dfrac{{52 \times 51 \times 50!}}{{2!50!}} = 51 \times 26$
Now the cards should be of different suits and different denominations. We know that there are $4$ suits (heart, club, spade, and diamond) with $13$ cards each. Suppose one card is chosen from a suit.
So the number of ways to select $1$ card from $13$ cards=${}^{13}{C_1}$
Now there are still $51$ cards left but we cannot select the same suit so $51 - 12 = 39$
And from the remaining card three cards will be of same denomination so the number of cards to choose from=$39 - 3 = 36$
The number of ways to select one card from the remaining $36$ cards=${}^{36}{C_1}$
Then the required ways to select two cards=${}^{13}{C_1}$ ×${}^{36}{C_1}$
We know that ${}^n{C_r} = \dfrac{{n!}}{{r!n - r!}}$
Then we get,
$ \Rightarrow {}^{13}{C_1} \times {}^{36}{C_1} = \dfrac{{13!}}{{1!13 - 1!}} \times \dfrac{{36!}}{{1!36 - 1!}}$
On solving we get,
$ \Rightarrow {}^{13}{C_1} \times {}^{36}{C_1} = \dfrac{{13 \times 12!}}{{12!}} \times \dfrac{{36 \times 35!}}{{35!}} = 13 \times 36$
Now the probability of drawing two cards of different suit and denomination will be
$ \Rightarrow $ P=$\dfrac{{{\text{required ways to select 2 cards}}}}{{{\text{total ways to select cards}}}}$
On putting the values we get,
$ \Rightarrow $ P=$\dfrac{{13 \times 36}}{{26 \times 51}}$
On solving we get,
$ \Rightarrow $ P=$\dfrac{{36}}{{2 \times 51}} = \dfrac{6}{{17}}$
Answer- The correct answer is D.

Note: Different denominations mean the number drawn on the card should be different like if the ace of one suit is already drawn then the next card cannot be an ace (of any suit). The student may go wrong if he/she also counts the other card from the pack of $51$ cards without subtracting the number of cards of the suit already drawn out. As then the cards will not be of different suites as asked in the question. Also here the student may get confused about why we have calculated the sample space as the total number of ways to select two cards. It is because we have to select two cards out of $52$cards of any denomination and suit as ample space means the collection of all possible outcomes.