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

$$\begin{gathered} \text{Given:\hspace{0.17em}Bag}=\left(7W+5B+4R\right)\mathrm{balls} \\ P\left(\text{atleast 3 balls are black}\right)=P\left(\text{exactly 3 black}\right)+P\left(\text{all 4 black}\right) \\ =\left(\frac{11}{16}\times \frac{5}{15}\times \frac{4}{14}\times \frac{3}{13}\times 4\right)+\left(\frac{5}{16}\times \frac{4}{15}\times \frac{3}{14}\times \frac{2}{13}\right) \\ =\frac{11}{14\times 13}+\frac{1}{2\times 14\times 13} \\ =\frac{22+1}{364} \\ =\frac{23}{364} \end{gathered}$$