Q. A bag contains 5 white and 3 black balls and 4 balls are successively drawn out and not replaced. The probability that they are alternately of different colours, is
A) $\frac{1}{196}$
B) $\frac{2}{7}$
C) $\frac{13}{56}$
D) $\frac{1}{7}$
Explanation
(d)
Required probability $= P(WBWB) + P(BWBW)$
$= \left( \ \frac{{\ ^{5}C}_{1}{{\times \ }^{3}C}_{1} \times {\ ^{4}C}_{1}{{\ \times}^{2}C}_{1}}{{\ ^{8}C}_{1}{\times \ ^{7}C}_{1}{{\ \times}^{6}C}_{1} \times {\ ^{5}C}_{1}} \right) + \left( \ \frac{{\ ^{5}C}_{1}{\times \ ^{3}C}_{1} \times {\ ^{4}C}_{1}{\times \ ^{2}C}_{1}}{{\ ^{8}C}_{1}{{\ \times}^{7}C}_{1} \times {\ ^{6}C}_{1} \times {\ ^{5}C}_{1}} \right)$
$= \frac{1}{14} + \frac{1}{14} = \frac{2}{14} = \frac{1}{7}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.