Question

A bag contains tickets numbered from $1$ to $20$. Two tickets are drawn. The probability that both the numbers are prime is
$1)$ $\dfrac{14}{95}$
$2)$ $\dfrac{7}{95}$
$3)$ $\dfrac{1}{95}$
$4)$ None of these

Answer 1

Hint: Since, in the question, we have given some data. So, we will calculate the required probability by using the formula of finding the number of ways that is $^{n}{{C}_{r}}$ , where $n$ denotes total number and $r$ denotes required number. Firstly, we will calculate the total number of possible ways with help of formula. Then we will calculate a favourable number of possible ways of getting a prime number by using formula. And for getting the required probability, we will calculate the ratio of favourable ways and total ways.

Complete step by step solution:
Since, we have given that the bag contains a total $20$ tickets and we need to draw $2$ tickets. So, we will use the formula for getting the possible number of ways of drawing the $2$ tickets as:
$\Rightarrow {}^{20}{{C}_{2}}$
We can expand it as:
$\Rightarrow \dfrac{20!}{2!.\left( 20-2 \right)!}$
We will solve the denominator bracketed term as:
$\Rightarrow \dfrac{20!}{2!.18!}$
Here, we can write it as:
$\Rightarrow \dfrac{20\times 19\times 18!}{1\times 2\times 18!}$
Since, $18!$ is available in both numerator and denominator, it will be canceled out as:
$\Rightarrow \dfrac{20\times 19}{1\times 2}$
After solving above step, we will get:
$\Rightarrow 10\times 19$
After solving this formula for getting the total possible ways of drawing $2$ tickets, we will get the value as:
$\Rightarrow 190$
Now, we will do the calculation to get the favourable number of possible ways of getting a prime number. Since, from $1$ to $20$ , we have total $8$ prime numbers and we will draw $2$ prime tickets. Thus, the favourable ways is:
$\Rightarrow {}^{8}{{C}_{2}}$
Here, we will expand it as:
$\Rightarrow \dfrac{8!}{2!.\left( 8-2 \right)!}$
We will solve the denominator bracketed term as:
$\Rightarrow \dfrac{8!}{2!.6!}$
Here, we can writenumerator as:
$\Rightarrow \dfrac{8\times 7\times 6!}{1\times 2\times 6!}$
Since, $6!$ is available in both numerator and denominator, it will be canceled out as:
$\Rightarrow \dfrac{8\times 7}{1\times 2}$
After solving above step, we will get:
$\Rightarrow 4\times 7$
After solving the above step, we will have next step below as:
$\Rightarrow 28$
Now, we will get the required probability by taking the ratio of favourable number of ways of drawing two tickets and total number of possible ways of drawing two tickets as:
$\Rightarrow \dfrac{\text{Favourable Number of Ways}}{\text{Total Number of possible Ways}}$
Here, we will put the values of favourable number of ways and total number of possible ways respectively as:
$\Rightarrow \dfrac{28}{190}$
Since, $2$ is a common number in both numerator and denominator. So, we will cancel out it and will have the above ratio as:
$\Rightarrow \dfrac{14}{95}$
Hence, this is the required probability as $\dfrac{14}{95}$.

Note: Since, the question is about the selection of the ball that means this type of question will follow the rule of possible ways of selection that comes under the rule of combination and will use the formula ${}^{n}{{C}_{r}}$ . Further, we can expand it as:
$\Rightarrow {}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$