Question

A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?
   (A) $\dfrac{10}{21}$
   (B) $\dfrac{11}{21}$
   (C) $\dfrac{2}{7}$
   (D) $\dfrac{5}{7}$

Answer 1

Hint: We will first find the number of ways by which we can select two balls from 7 balls. In the second step we will find the number of ways by which we can select two balls from 5 balls. Final probability can be calculated by: $\dfrac{\text{Favoured Outcome}}{\text{Total Outcome}}$.

Complete step-by-step solution -
It is given in the question that in a bag there are 2 red, 3 green and 2 blue balls. Two balls are drawn at random then we have to find the probability that the ball drawn is not blue. Now, total number of balls in the bag = $2+3+2=7balls$. Then the number of ways of finding two balls out of seven balls will be ${}^{7}{{c}_{2}}$.

 We know that ${}^{n}{{c}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$, therefore ${}^{7}{{c}_{2}}=\dfrac{7!}{2!\left( 7-2 \right)!}$ = $\dfrac{7\times 6\times 5!}{2!\times 5!}$. Cancelling $5!$ from the numerator and denominator, we get $\dfrac{7\times 6}{2}=21$. So, there are 21 ways of picking 2 balls at random from the bag containing 7 balls.

Now, we have to find that the ball drawn will not be blue so our sample will be \[total\text{ }balls\text{ }\text{ }number\text{ }of\text{ }blue\text{ }balls\] = $7-2=5balls$. We will find the number of ways of picking 2 balls from 5 balls as we are not considering blue balls in our sample. Therefore, the number of ways of finding two balls which are not blue will be ${}^{5}{{c}_{2}}$. Using the formula ${}^{n}{{c}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$, we get ${}^{5}{{c}_{2}}=\dfrac{5!}{2!\left( 5-2 \right)!}$ = $\dfrac{5\times 4\times 3!}{2!\times 3!}$. On cancelling $3!$ from both numerator and denominator, we get $\dfrac{5\times 4}{2}=10$. Therefore, there are 10 ways of picking 2 balls at random from the bag containing 5 non-blue balls.

Now, the probability that the 2 balls drawn are not blue is given by $\dfrac{\text{Favoured Outcome}}{\text{Total Outcome}}$. We have favourable outcomes as 10 and total outcomes as 21. Therefore, $\text {Required}\ \text{Probability}=\dfrac{10}{21}$, hence option a) is correct.

Note: Student may confuse and they find non-favoured outcomes instead of favoured outcomes, they calculate non-favoured outcome as ${}^{2}{{c}_{2}}$ because there are 2 blue balls in bag and we have to draw 2 balls from bag which are not blue, but to find the probability that the ball drawn are not blue is given by $\dfrac{{}^{5}{{c}_{2}}}{{}^{7}{{c}_{2}}}=\dfrac{10}{21}$.