A bag contains 5 distinct Red, 4 distinct Green and 3 distinct Black balls. Balls are draw

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14.Probability

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A bag contains $5$ distinct Red, $4$ distinct Green and $3$ distinct Black balls. Balls are drawn one by one without replacement,then the probability of getting a particular red ball in fourth draw is-

A

$\frac{1}{12}$

B

$\frac{223}{1188}$

C

$\frac{335}{1320}$

D

$\frac{5}{12}$

Solution

Required probability $=\frac{11}{12} \cdot \frac{10}{11} \cdot \frac{9}{10} \cdot \frac{1}{9}=\frac{1}{12}$

Standard 11

Mathematics

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