# 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??

Arun Kumar IIT Delhi
8 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
$P(E_{1} \cap E_{2}) = P(RBR) + P(RRB) = \frac{5}{8} \times \frac{3}{7} \times \frac{4}{6} + \frac{5}{8} \times \frac{4}{7} \times \frac{3}{6} = \frac{120}{336} P(E_{2}/E_{1}) = \frac{P(E_{1} \cap E_{2})}{P(E_{1})} = \frac{\frac{120}{336}}{\frac{210}{336}} = \frac{4}{7}$