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. $\frac{10}{21}$
• B. $\frac{11}{21}$
• C. $\frac{2}{7}$
• D. $\frac{5}{7}$

Answer 1

The correct option is A $\frac{10}{21}$

Total no of chances to draw $2$ balls at random from $7$ coloured balls,$n\left(s\right)=7_{C_2}=\frac{7!}{5!2!}=\frac{7\times 6\times 5!}{5!\times 2}=21$

let $E$ be an event to draw $2$ balls other than blue .

Number of chances to draw two balls other than blue are,

$n\left(E\right)=2_{C_2}+3_{C_2}+2_{C_1}.3_{C_1}$

$=\frac{2!}{0!2!}+\frac{3!}{1!2!}+\frac{2!}{1!1!}.\frac{3!}{1!2!}=1+3+6=10$

the probability that none of balls drawn is blue is,$p\left(E\right)=\frac{n\left(E\right)}{n\left(s\right)}=\frac{10}{21}$