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 ?

Engineering

Mathematics

Basic Definitions Related to Probability

Question

$\frac{10}{21}$

$\frac{11}{21}$

$\frac{2}{7}$

$\frac{5}{7}$

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Total no of chances to draw 2 balls at random from 7 coloured balls,

n (s) = 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 (E) = 2_{C_{2}} + 3_{C_{2}} + 2_{C_{1}} \cdot 3_{C_{1}}

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

the probability that none of balls drawn is blue is,

P (E) = \frac{n (E)}{n (s)} = \frac{10}{21}

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