Question

A bag contains 5 white, 7 red and 3 black balls. If three balls are drawn one by one without replacement, find the probability that none is red.

Answer 1

Consider the given events.
A = A white or black ball in the first draw
B = A white or black ball in the second draw
C = A white or black ball in the third draw

$$\begin{gathered} \mathrm{Now}, \\ P\left(A\right)=\frac{8}{15} \\ P\left(B/A\right)=\frac{7}{14}=\frac{1}{2} \\ P\left(C/A\cap B\right)=\frac{6}{13} \\ \therefore \mathrm{Required}\mathrm{probability}=P\left(A\cap B\cap C\right) \\ =P\left(A\right)\times P\left(B/A\right)\times P\left(C/A\cap B\right) \\ =\frac{8}{15}\times \frac{1}{2}\times \frac{6}{13} \\ =\frac{8}{65} \end{gathered}$$