A bag contains 10 similar balls,of which 4 ar | vedclass

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(B) Total number of balls $= 10$. Number of red balls $= 6$. We draw $3$ balls one after the other without replacement. The probability of drawing the

Vedclass · Accepted Answer

$\frac{1}{6}$

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(B) Total number of balls $= 10$. Number of red balls $= 6$. We draw $3$ balls one after the other without replacement. The probability of drawing the first red ball is $P(R_1) = \frac{6}{10}$. The probability of drawing the second red ball given the first was red is $P(R_2|R_1) = \frac{5}{9}$. The probability of drawing the third red ball given the first two were red is $P(R_3|R_1 \cap R_2) = \frac{4}{8}$. The required probability is $P(R_1 \cap R_2 \cap R_3) = \frac{6}{10} 	imes \frac{5}{9} 	imes \frac{4}{8} = \frac{120}{720} = \frac{1}{6}$.

$A$ bag contains $10$ similar balls,of which $4$ are blue and $6$ are red. Three balls are taken out at random from the bag one after the other without replacement. The probability that all the three balls drawn are red is

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