Question

A bag contains $7$ white, $5$ black and $4$ red balls. Four balls are drawn without replacement. Find the probability that at least three balls are black.

Answer 1

We have,

$BagContains=(7W+5B+4R)balls$

$P\left(atleast3ballsareblack\right)=P\left(exactly3black\right)+P\left(all4black\right)$

$=(\frac{11}{16}\times \frac{5}{15}\times \frac{4}{14}\times \frac{3}{13}\times 4)+(\frac{5}{16}\times \frac{4}{15}\times \frac{3}{14}\times \frac{2}{13})$

$=(11\times \frac{1}{14}\times \frac{1}{13})+(\frac{1}{4}\times \frac{1}{7}\times \frac{1}{13})$

$=\left(\frac{11}{182}\right)+\left(\frac{1}{364}\right)$

$=\frac{22+1}{364}$

$=\frac{23}{364}$

Hence, this is the answer.