Question

A bag contains $4$ white, $6$ red and $7$ black balls. If four balls are drawn one by one without replacement, then the probability that all balls are black is

• A. $\frac{1}{68}$
• B. $\frac{1}{34}$
• C. $\frac{1}{136}$
• D. $\frac{2}{65}$

Answer 1

The correct option is A $\frac{1}{68}$

Getting black ball in first draw $=\frac{7}{17}$
Getting black ball in second draw $=\frac{6}{16}$
Getting black ball in third draw $=\frac{5}{15}$
Getting black ball in fourth draw $=\frac{4}{14}$
So, required probability $=\frac{7}{17}\times \frac{6}{16}\times \frac{5}{15}\times \frac{4}{14}=\frac{1}{68}$