A bag contains 3 red and 7 black balls. Two b | vedclass

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(B) Total number of balls in the bag = $3 	ext{ (red)} + 7 	ext{ (black)} = 10 	ext{ balls}$.Since the first ball taken out is red,it is not replac

Vedclass · Accepted Answer

$\frac{2}{9}$

Vedclass · Answer

(B) Total number of balls in the bag = $3 	ext{ (red)} + 7 	ext{ (black)} = 10 	ext{ balls}$.Since the first ball taken out is red,it is not replaced.After removing one red ball,the remaining number of balls in the bag = $10 - 1 = 9$.The remaining number of red balls in the bag = $3 - 1 = 2$.Therefore,the probability that the second ball is red,given that the first ball was red,is the number of remaining red balls divided by the total number of remaining balls.Probability = $\frac{2}{9}$.

$A$ bag contains $3$ red and $7$ black balls. Two balls are taken out at random,without replacement. If the first ball taken out is red,what is the probability that the second ball taken out is also red?

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