| A basket contains 6 red, 5 green and 8 blue balls. If four balls are picked at random, what is the probability that all four of them are either red or any two out of the four are green? |
A) \[\frac{5}{1292}\]
B) \[\frac{925}{3876}\]
C) \[\frac{359}{1938}\]
D) \[\frac{11}{3876}\]
Correct Answer: B
Solution :
| Total number of balls \[=6+5+8=19\] |
| Number of ways of selecting 4 balls out of \[19={{\,}^{19}}\,{{C}_{4}}\] |
| \[=\frac{19\times 18\times 17\times 16}{1\times 2\times 3\times 4}=3876\] |
| total number of favorable cases |
| \[={}^{6}{{C}_{4}}+{}^{5}{{C}_{2}}\times {}^{14}{{C}_{2}}\] |
| \[=\frac{6\times 5\times 4\times 3}{1\times 2\times 3\times 4}+\frac{5\times 4}{1\times 2}\times \frac{14\times 13}{1\times 12}\] |
| \[=15+(10\times 7\times 13)\] |
| \[=15+910=925\] |
| \[\therefore \]Required probability \[=\frac{925}{3876}\] |
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