Probability Part 3

Concept:Bayes' theorem is used to find the probability of a cause (number of red balls) given an observed outcome (three black balls drawn).Explanation:Let $E_k$ be the event that the bag contains $k$ red balls, where $k = 0,1,\dots,10$.All 11 compositions are equally likely, so $P(E_k) = \frac{1}{11}$ for each $k$.Let $A$ be the event that all three drawn balls are black.Given $k$ red balls, number of black balls = $10 - k$.Probability of drawing 3 black balls from these is $P(A|E_k) = \frac{\binom{10-k}{3}}{\binom{10}{3}}$, valid only when $10-k \ge 3$ i.e. $k \le 7$.We need $P(E_1|A)$, the probability that $k = 1$ (1 red, 9 black) given $A$.By Bayes' theorem: $P(E_1|A) = \frac{P(E_1) \cdot P(A|E_1)}{\sum_{k=0}^{7} P(E_k) \cdot P(A|E_k)}$.Since $P(E_k)$ is constant, it cancels: $P(E_1|A) = \frac{\binom{9}{3}}{\binom{10}{3} + \binom{9}{3} + \cdots + \binom{3}{3}}$.Numerator: $\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$.Denominator sum uses the identity $\binom{3}{3} + \binom{4}{3} + \cdots + \binom{10}{3} = \binom{11}{4} = 330$.Therefore $P(E_1|A) = \frac{84}{330} = \frac{14}{55}$.Answer:$\frac{14}{55}$, which corresponds to option D.