A bag contains 3 white, 3 black and 2 red balls. One by one, three balls a

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A bag contains 3 white, 3 black and 2 red balls. One by one, three balls are drawn without replacing them. Then the probability that the third ball is red , is given by

Options:

A . 524

B . 112

C . 14

D . 12

Answer: Option C
:
C
Number of white balls = 3, black balls = 3 red balls = 2
Since drawn balls are not replaced, for third ball to red, we have following patterns:

E_{1} = W W R, B B R, E_{2} = B W R, W B R; a n d E_{3} = R B R, R W R, B R R, W R R

P (E_{1}) = 2 \times \frac{3}{8} \times \frac{2}{7} \times \frac{2}{6}, P (E_{2}) = 2 \times \frac{3}{8} \times \frac{3}{7} \times \frac{2}{6}, P (E_{3}) = \frac{4 \times 2 \times 3 \times 1}{8.7.6}

Required probability

= P (E_{1}) + P (E_{2}) + P (E_{3}) = \frac{1}{4}

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