| 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) \[\frac{1}{7}\]
Correct Answer: D
Solution :
| Required probability =P(WBWB) + P(BWBW) |
| \[=\,\,\left( \frac{{}^{3}{{C}_{1}}\times {}^{5}{{C}_{1}}\times {}^{2}{{C}_{1}}\times {}^{4}{{C}_{1}}}{{}^{8}{{C}_{1}}\times {}^{7}{{C}_{1}}\times {}^{6}{{C}_{1}}\times {}^{5}{{C}_{1}}} \right)+\left( \frac{{}^{3}{{C}_{1}}\times {}^{5}{{C}_{1}}\times {}^{2}{{C}_{1}}\times {}^{4}{{C}_{1}}}{{}^{8}{{C}_{1}}\times {}^{7}{{C}_{1}}\times {}^{6}{{C}_{1}}\times {}^{5}{{C}_{1}}} \right)\]\[=\,\,\frac{1}{14}+\frac{1}{14}=\frac{2}{14}=\frac{1}{7}\] |
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