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 number of balls = (2 + 3 + 2) = 7
Let S be the sample space.
Then, n(S) = Number of ways of drawing 2 balls out of 7
${=}^7{C}_2$
$=\frac{(7\times 6)}{(2\times 1)}$
$=21$
Let E = Event of drawing 2 balls, none of which is blue.
$\therefore n\left(E\right)=$ Number of ways of drawing 2 balls out of (2 + 3) balls.
${=}^5{C}_2$
$=\frac{(5\times 4)}{(2\times 1)}$
$=10$
$\therefore P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{10}{21}$