Probability Part 3

Concept: This is a problem of conditional probability, requiring the application of Bayes' Theorem. Since the prior probability of the initial composition ($k$) is unknown, we assume a uniform prior distribution over all possible states that allow the observed event to occur.Formula:$P(H_1 \mid E) = \frac{P(E \mid H_1) P(H_1)}{\sum_{k} P(E \mid H_k) P(H_k)}$Where $H_k$ is the hypothesis that the bag has $k$ red balls, $E$ is the event that 3 black balls are drawn, and $P(E | H_k) = \frac{\binom{10-k}{3}}{\binom{10}{3}}$.Solution:Let $N=10$ be the total number of balls. Let $H_k$ be the hypothesis that the bag contains $k$ red balls and $10-k$ black balls.The event $E$ is drawing 3 black balls.The probability of $E$ given $H_k$ is:$P(E | H_k) = \frac{\binom{10-k}{3}}{\binom{10}{3}}$Since no prior information is given, we assume a uniform prior $P(H_k) = C$ for all possible states $k$. The event $E$ requires at least 3 black balls, so $10-k \geq 3$, meaning $k \leq 7$. Thus, $k$ can range from $0$ to $7$.Using Bayes' Theorem, the desired probability is proportional to $P(E | H_k)$:$P(H_1 | E) = \frac{P(E | H_1)}{\sum_{k=0}^{7} P(E | H_k)}$1. Calculate the numerator (for $k=1$):$P(E | H_1) = \frac{\binom{10-1}{3}}{\binom{10}{3}} = \frac{\binom{9}{3}}{120}$$\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$$P(E | H_1) = \frac{84}{120}$2. Calculate the denominator (Sum of probabilities):$\sum\limits_{k=0}^{7} P(E | H_k) = \frac{1}{120} \sum\limits_{k=0}^{7} \binom{10-k}{3}$Let $j = 10-k$. As $k$ goes from $0$ to $7$, $j$ goes from $10$ to $3$.$\sum\limits_{k=0}^{7} \binom{10-k}{3} = \sum\limits_{j=3}^{10} \binom{j}{3}$Using the Hockey-stick identity $\sum\limits_{i=r}^{n} \binom{i}{r} = \binom{n+1}{r+1}$:$\sum\limits_{j=3}^{10} \binom{j}{3} = \binom{10+1}{3+1} = \binom{11}{4}$$\binom{11}{4} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 11 \times 10 \times 3 = 330$$\sum\limits_{k=0}^{7} P(E | H_k) = \frac{330}{120}$3. Find the conditional probability:$P(H_1 | E) = \frac{\frac{84}{120}}{\frac{330}{120}} = \frac{84}{330}$Simplifying the fraction by dividing numerator and denominator by 6:$P(H_1 | E) = \frac{14}{55}$Shortcut:The probability is the ratio of the number of ways to draw 3 black balls if the bag has 1 red ball ($\binom{9}{3}$) to the total number of ways to draw 3 black balls across all possible compositions ($\sum\limits_{j=3}^{10} \binom{j}{3} = \binom{11}{4}$).$P = \frac{\binom{9}{3}}{\binom{11}{4}} = \frac{84}{330} = \frac{14}{55}$Answer: D. $\frac{14}{55}$