A) \[\frac{15}{286}\]
B) \[\frac{7}{286}\]
C) \[\frac{105}{286}\]
D) \[\frac{35}{286}\]
Correct Answer: A
Solution :
Required probability = Probability of getting exactly two white balls and four black balls is 1 to 6th draw(X) probability of getting third white ball in 7th draw. \[=\frac{^{3}{{C}_{2}}{{\times }^{10}}{{C}_{4}}}{^{13}{{C}_{6}}}\times \frac{1}{7}=\frac{15}{286}\]
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