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A bag contains 5 red nd 3 blue balls. if 3 balls are drawn at random without replacement the probability that exactly two of three balls were red, the first ball being red is??

A bag contains 5 red nd 3 blue balls. if 3 balls are drawn at random without replacement the probability that exactly two of three balls were red, the first ball being red is??

Grade:12th pass

2 Answers

Arun Kumar IIT Delhi
askIITians Faculty 256 Points
9 years ago
Hello Student,
Total no of ways of selecting 3 balls out of total 8 balls=8C3
Total no of selecting 2 red ball out of 5 red ball and 1 blue ball out of 3 blue balls=5C2*3C1
Hence the probability of exactly 2 red balls out of 3 balls=(5C2*3C1)/8C3
=(10)(3)/56=30/56=15/28
Thanks & Regards
Arun Kumar
Btech, IIT Delhi
Askiitians Faculty
Navjot SIngh
11 Points
6 years ago
Your answer is wrong since the question is of conditional probability. It is about finding the probability that exactly two will be red, given that the first ball is red. 
Let E1= Event that first ball being red And E2= Event that exactly two of three balls being red
P(E1) = P(RRR) + P(RRB) + P(RBR) + P(RBB) = 5/8 x 4/7 x 3/6 + 5/8 x 4/7 x 3/6 + 5/8 x 3/7 x 4/6 + 5/8 x 3/7 x 2/6 = 210/336
P(E_{1} \cap E_{2}) = P(RBR) + P(RRB) = \frac{5}{8} \times \frac{3}{7} \times \frac{4}{6} + \frac{5}{8} \times \frac{4}{7} \times \frac{3}{6} = \frac{120}{336} P(E_{2}/E_{1}) = \frac{P(E_{1} \cap E_{2})}{P(E_{1})} = \frac{\frac{120}{336}}{\frac{210}{336}} = \frac{4}{7}

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