Question

Two numbers are selected randomly from the set $S=\left\{ 1,2,3,4,5,6 \right\}$ without replacement one by one the probability that the minimum of the two numbers is less than $4$ is
(A)$\dfrac{1}{15}$
(B)$\dfrac{14}{15}$
(C)$\dfrac{1}{5}$
(D)$\dfrac{4}{5}$

Answer 1

Hint: To solve this question we need to know the concept of arrangements which is combination. The first step in the calculation of the problem is to find the total number of ways the two numbers can be selected. The next step is to find the ways in which numbers selected have a minimum number less than $4$. Then comes the last step which is to find the probability.

Complete step-by-step solution:
The question asks us to find the probability, when two numbers are selected randomly from a set which contains numbers from $1$ to $6$ where in the number is selected without any replacement. This has a condition that any of the numbers selected the minimum should be less than $6$ .
Number of ways two numbers can be selected from a set of 6 numbers is ${}^{6}{{C}_{2}}$.
Now the number ${}^{6}{{C}_{2}}$ will be multiplied to $2!$ because the number selected may be firstly choosen or not.
Number of ways the two numbers can be selected from set of $6$ numbers
 $\Rightarrow 2{}^{6}{{C}_{2}}$
On calculating the above combination we get:
$\Rightarrow 2\times \dfrac{6!}{2!\left( 6-2 \right)!}$
$\Rightarrow 2\times \dfrac{6!}{2!4!}$
$\Rightarrow 2\times \dfrac{6\times 5\times 4!}{2!4!}$
$\Rightarrow 6\times 5$
\[\Rightarrow 30\]ways
Number of ways in which the minimum of the two numbers selected from the set is not less than $4$ is when the minimum number selected has a minimum number as $4$ or 5 or $6$. The selections which do not have a minimum less than $4$ are $\text{4,4 or 4,5 or 5,5 or 5,6 or 4,6 or 6,6}$ which is $6$ ways.
The number of ways the minimum of the two numbers is not less than $4$ = $6$ways
The number of ways the minimum of the two numbers is less than $4$= Total ways – Number of ways number is not less than $4$
The number of ways the minimum of the two numbers is less than$4$= $30-6$
$\Rightarrow 24$ways
$\text{Probability = }\dfrac{\text{Favourable Ways}}{\text{Total Ways}}$
In this question favourable ways is that the minimum of the two numbers is less than $4$which is $24$ and total ways are$30$ .
$\Rightarrow \dfrac{24}{30}$
On converting the fraction into its lowest term we get:
$\Rightarrow \dfrac{4}{5}$
$\therefore $ The probability that the minimum of the two numbers is less than $4$when $2$ numbers are selected randomly from the set of $6$ numbers is $\left( D \right)\dfrac{4}{5}$ .

Note: We should know about combinations which is a mathematical technique that determines the number of possible arrangements in a collection. It is represented as ${}^{a}{{C}_{b}}$ here $a$ is the total item and $b$ is the number of item on which condition need to be applied whereas $C$ is to denote combination. ${}^{a}{{C}_{b}}=\dfrac{a!}{b!\left( a-b \right)!}$ , where $''!''$ is the sign for factorial.