ProbabilityHard
Question
A bag contains 10 balls out of which k are red and $(10 - k)$ are black, where $0 \leq k \leq 10$. If three balls are drawn at random without replacement and all of them are found to be black, then the probability that the bag contains 1 red and 9 black balls is :
Options
A.$\frac{7}{11}$
B.$\frac{7}{55}$
C.$\frac{7}{110}$
D.$\frac{14}{55}$
Solution
Probability $= \frac{\ ^{1}C_{0} \cdot \ ^{9}C_{3}}{\sum_{k = 0}^{10}\mspace{2mu}\mspace{2mu}\ ^{k}C_{0} \cdot \ ^{10 - k}C_{3}}$
$$\begin{matrix} & \ = \frac{\ ^{9}C_{3}}{\ ^{10}C_{3} + \ ^{9}C_{3} + \ ^{8}C_{3} + \ldots \cdot \ ^{3}C_{3}} \\ & \ = \frac{\ ^{9}C_{3}}{\ ^{11}C_{4}} = \frac{14}{55} \end{matrix}$$
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