Question

A card is drawn at random from a well shuffled pack of $52$ cards. The probability of getting a two of heart or two of diamonds is
1)$$$\dfrac{1}{{26}}$ 2)\dfrac{2}{{17}}$ $$3)\dfrac{1}{{13}}$
4)$$$$None{\text{ }}of{\text{ }}these

Answer 1

Hint : We have to find the probability of getting two of hearts or a two of diamonds . We solve this question using the knowledge of cards and also the concept of permutation and combination . We first find the total possible arrangements and also the favourable outcomes using the formula of combination . The probability is given by favourable outcomes to the total possible arrangements .

Complete step-by-step answer :
Given :
Total number of cards $= {\text{ }}52$
Number of cards drawn $= {\text{ }}1$
Total number of favourable outcome = a two of heart or a two of black
( as in a deck of cards there is only one two of heart and only one two of diamond )
Total number of favourable outcome $= {\text{ }}1{\text{ }} + {\text{ }}1$
Total number of favourable outcome$= {\text{ }}2$
We have to choose only one card from the deck
So ,
The favourable outcomes $= {\text{ }}{}^2{C_1}$
Total possible outcome $= {\text{ }}{}^{52}{C_1}$
Now ,
The probability of getting a two of heart or a two of diamond $= \dfrac{{{\text{ }}{}^2{C_1}{\text{ }}}}{{{\text{ }}{}^{52}{C_1}}}$
Using the formula of combination
${}^n{C_r} = {\text{ }}\dfrac{{n!}}{{r!{\text{ }} \times {\text{ }}\left( {n{\text{ }} - {\text{ }}r} \right)!}}$
On solving , we get
The required probability $= {\text{ }}\dfrac{2}{{52}}$
After cancelling the term
The required probability $= {\text{ }}\dfrac{1}{{26}}$
Hence , the required probability of getting a two of heart or two of diamonds is$\dfrac{1}{{26}}$ .
Thus , the correct option is $\left( 1 \right)$
So, the correct answer is “Option 1”.

Note : Corresponding to each combination of ${}^n{C_r}$ we have $r!$ permutations, because $r$ objects in every combination can be rearranged in $r!$ ways . Hence , the total number of permutations of $n$different things taken $r$ at a time is ${}^n{C_r} \times {\text{ }}r!$ . Thus $\;{}^n{P_r}{\text{ }} = {\text{ }}{}^n{C_r}{\text{ }} \times {\text{ }}r!{\text{ }},{\text{ }}0 < {\text{ }}r{\text{ }} \leqslant n$
Also , some formulas used :
${}^n{C_1} = {\text{ }}n$
${}^n{C_2}{\text{ }} = {\text{ }}\dfrac{{n\left( {n - 1} \right)}}{2}$
${}^n{C_0}{\text{ }} = {\text{ }}1$
${}^n{C_n} = {\text{ }}1$