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1. A basket contains 4 red, 5 blue and 3 green marbles. If 2 marbles are drawn at random from the basket, What is the probability that both are red ?

A. $\frac{3}{7}$

B. $\frac{1}{2}$

C. $\frac{1}{11}$

D. $\frac{1}{6}$

Solution:

Total number of balls = (4 + 5 + 3) = 12
Let E be the event of drawing 2 red balls.
Then, n (E) =

^{4} C_{2} = \frac{4 \times 3}{2 \times 1}

= 6
Also n (S) =

^{12} C_{2} = \frac{12 \times 11}{2 \times 1}

= 66

∴ P (E) = \frac{n (E)}{n (S)} = \frac{6}{66} = \frac{1}{11}

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