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) $\frac{1}{7}$
D) $\frac{13}{56}$
Explanation
(c)
$\therefore$ Required probability
$= P(WBWB) + (BWBW)$
$= \frac{\ ^{5}C_{1}{\ \times}^{3}C_{1}{\times \ }^{4}C_{1}{\times \ }^{2}C_{1}}{\ ^{8}C_{1}{\times \ }^{7}C_{1} \times \ ^{6}C_{1}{\times \ }^{5}C_{1}} + \frac{\ ^{3}C_{1} \times \ ^{5}C_{1}{\times \ }^{2}C_{1}{\times \ }^{4}C_{1}}{\ ^{8}C_{1}{\ \times}^{7}C_{1}{\times \ }^{6}C_{1}{\times \ }^{5}C_{1}}$
$= 2\left( \frac{5 \times 3 \times 4 \times 2}{8 \times 7 \times 6 \times 5} \right) = \frac{1}{7}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.