Question

If $2$ cards are drawn from a well shuffled pack of 52 cards, the probability that they are of the same colours without replacement is:
(a) $\left( \dfrac{1}{2} \right)$
(b) $\left( \dfrac{26}{51} \right)$
(c) $\left( \dfrac{25}{51} \right)$
(d) $\left( \dfrac{25}{102} \right)$

Answer 1

Hint: In a pack of $52$ cards, there are $26$ red cards and $26$ black cards. We have to choose $2$ cards from these $52$ cards such that the $2$ cards are of the same colour. So, we can use the concept of permutation and combination to find the number of ways in which we can select $2$ cards from a pack of $52$ cards such that the $2$ cards are of the same colour.

Complete step-by-step answer:
Before proceeding with the question, we must know that there are $26$ red cards and $26$ black cards in a pack of $52$ cards.
The probability of an event $X$ is denoted by $P\left( X \right)$ and it is given by the formula,
$P\left( X \right)=$ ratio of number of outcomes that are favourable to event $X$ and the total number of outcomes that are possible $..........\left( 1 \right)$
In this question, the number of favourable outcomes is equal to the number of ways in which we can draw two cards of the same colour. This means that we either select two cards out from $26$ red cards or we select two cards from $26$ black cards.
From the concept of permutation and combination, the number of ways in which we can select two red cards out of $26$ red cards is equal to ${}^{26}{{C}_{2}}$ .
Similarly, from the concept of permutation and combination, the number of ways in which we can select two black cards out of $26$ black cards is equal to ${}^{26}{{C}_{2}}$ .
Since either of the above two cases are possible, the number of outcomes that are favourable can be calculated by simply adding the two numbers we got in the above two cases.
$\Rightarrow$ Number of favourable outcomes $={}^{26}{{C}_{2}}+{}^{26}{{C}_{2}}$
$\Rightarrow$ Number of favourable outcomes $=2\times {}^{26}{{C}_{2}}...........\left( 2 \right)$.
To calculate the total number of possible outcomes, we will calculate the number of ways in which we can select two cards from a pack of $52$ cards containing both red and black cards. So, using the concept of permutation and combination, the number of ways in which we can select two cards out of the pack of $52$ card is equal to ${}^{52}{{C}_{2}}.........\left( 3 \right)$.
Substituting the number of favourable outcomes from equation $\left( 2 \right)$ and the total number of outcomes possible from equation $\left( 3 \right)$ in formula $\left( 1 \right)$, we get,
$P\left( X \right)=\dfrac{2\times {}^{26}{{C}_{2}}}{{}^{52}{{C}_{2}}}$
We have a formula ${}^{n}{{C}_{r}}=\dfrac{\left( n \right)!}{r!\left( n-r \right)!}$ . Using this formula, we get,
$\begin{aligned} & P\left( X \right)=\dfrac{2\times \dfrac{\left( 26 \right)!}{2!\left( 26-2 \right)!}}{\dfrac{\left( 52 \right)!}{2!\left( 52-2 \right)!}} \\\ & \Rightarrow P\left( X \right)=\dfrac{2\times \dfrac{\left( 26 \right)\left( 25 \right)\left( 24 \right)!}{2!\left( 24 \right)!}}{\dfrac{\left( 52 \right)\left( 51 \right)\left( 50 \right)!}{2!\left( 50 \right)!}} \\\ & \Rightarrow P\left( X \right)=\dfrac{2\times \left( 26 \right)\left( 25 \right)}{\left( 52 \right)\left( 51 \right)} \\\ & \Rightarrow P\left( X \right)=\dfrac{25}{51} \\\ \end{aligned}$

Therefore the correct answer is option (c)

Note: There is a possibility that one may commit mistake by writing the number of favourable cases equal to ${}^{26}{{C}_{2}}\times {}^{26}{{C}_{2}}$ instead of ${}^{26}{{C}_{2}}+{}^{26}{{C}_{2}}$. But as we have to choose two cards either from red or black cards, the number of favourable cases will be equal to ${}^{26}{{C}_{2}}+{}^{26}{{C}_{2}}$.