Answer to Question #199299 in Statistics and Probability for MARIA NDEEM

Question

We have 3 blue and 5 Red balls lay in a box, three balls are chosen randomly out of the box find the probability distribution for number of blue balls without replacement.

Answer 1

B = Blue ball

R = Red ball

Let X denotes the event of number of blue balls are chosen out of three balls drawn from the box.

X = 0 (RRR)

$P(X=0) = \frac{5}{8} \times \frac{4}{7} \times \frac{3}{6}=\frac{60}{336}=0.178$

X = 1 (BRR, RBR, RRB)

$P(X=1) = (\frac{3}{8} \times \frac{5}{7} \times \frac{4}{6})+(\frac{5}{8} \times \frac{3}{7} \times \frac{4}{6})+(\frac{5}{8} \times \frac{4}{7} \times \frac{3}{6})= \frac{180}{336}=0.535$

X=2 (BBR, BRB, RBB)

$P(X=2) = (\frac{3}{8} \times \frac{2}{7} \times \frac{5}{6})+(\frac{3}{8} \times \frac{5}{7} \times \frac{2}{6})+(\frac{5}{8} \times \frac{3}{7} \times \frac{2}{6}) = \frac{90}{336}=0.267$

X=3 (BBB)

$P(X=3) = \frac{3}{8} \times \frac{2}{7} \times \frac{1}{6} = \frac{6}{336} = 0.017$

Probability distribution: