Algebra> A bag contains 5 red nd 3 blue balls. if ...

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??

Miss Pooja , 10 Years ago

Grade 12th pass

2 Answers

Arun Kumar

Last Activity: 10 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

Last Activity: 7 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
$P(E_{1} \cap E_{2}) = P(RBR) + P(RRB) = \frac{5}{8} \times \frac{3}{7} \times \frac{4}{6} + \frac{5}{8} \times \frac{4}{7} \times \frac{3}{6} = \frac{120}{336} P(E_{2}/E_{1}) = \frac{P(E_{1} \cap E_{2})}{P(E_{1})} = \frac{\frac{120}{336}}{\frac{210}{336}} = \frac{4}{7}$