Question ID: #920
A bag contains 10 balls out of which k are red and $(10-k)$ are black, where $0\le k\le10.$ 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:
- (1) $\frac{7}{11}$
- (2) $\frac{7}{55}$
- (3) $\frac{7}{110}$
- (4) $\frac{14}{55}$
Solution:
Let $E$ be the event that 3 balls drawn are black.
Let $H_k$ be the hypothesis that the bag contains $k$ red balls (and thus $10-k$ black balls).
The possible values for $k$ are $0, 1, 2, …, 10$.
Assuming each initial composition is equally likely, $P(H_k) = \frac{1}{11}$.
The probability of drawing 3 black balls given $k$ red balls is:
$$P(E|H_k) = \frac{{}^{10-k}C_3}{{}^{10}C_3}$$
Note that for $E$ to happen, we need at least 3 black balls, so $10-k \ge 3 \Rightarrow k \le 7$. If $k > 7$, $P(E|H_k) = 0$.
We need to find $P(H_1|E)$ (Probability that bag contains 1 red ball, i.e., $k=1$).
Using Bayes’ Theorem:
$$P(H_1|E) = \frac{P(E|H_1)P(H_1)}{\sum_{k=0}^{10} P(E|H_k)P(H_k)}$$
Since $P(H_k)$ is constant, it cancels out:
$$P(H_1|E) = \frac{{}^{9}C_3}{\sum_{k=0}^{7} {}^{10-k}C_3}$$
Numerator:
$${}^{9}C_3 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$$
Denominator:
$$S = {}^{10}C_3 + {}^{9}C_3 + {}^{8}C_3 + … + {}^{3}C_3$$
Using the hockey-stick identity $\sum_{r=n}^{m} {}^rC_n = {}^{m+1}C_{n+1}$:
$$S = \sum_{r=3}^{10} {}^rC_3 = {}^{11}C_4$$
$${}^{11}C_4 = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 330$$
Required Probability:
$$P = \frac{84}{330}$$
Divide by 6:
$$P = \frac{14}{55}$$
Ans. (4)
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