A bag contains 1O 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:

Bayes theorem in combination problems

Bag has 10 balls, k red and (10-k) black with k uniform on {0,…,10}. Three balls drawn without replacement are all black.

First find P(3 black|k). Then apply Bayes theorem: P(k=1|3 black)=P(k=1 and 3 black)/P(3 black). Compute the numerator and denominator using combinations.

Step 1: Prior probability of k=1 is $$P(k=1)=\frac{1}{11}$$.
Step 2: Given k, number of black balls is $$(10-k)$$. Probability of drawing 3 black without replacement is
$$P(3\,black\mid k)=\frac{\displaystyle{\binom{10-k}{3}}}{\displaystyle{\binom{10}{3}}}.$$
Step 3: Compute numerator for k=1:
$$P(k=1,3\,black)=P(k=1)\times P(3\,black\mid k=1)=\frac{1}{11}\times\frac{\binom{9}{3}}{\binom{10}{3}}=\frac{1}{11}\times\frac{84}{120}=\frac{84}{1320}=\frac{7}{110}.$$
Step 4: Compute total probability of drawing 3 black:
$$P(3\,black)=\sum_{k=0}^{10}P(k)P(3\,black\mid k)=\frac{1}{11}\sum_{k=0}^{10}\frac{\binom{10-k}{3}}{\binom{10}{3}}. $$
Let $$i=10-k$$, then $$i$$ goes from 0 to 10. So
$$\sum_{k=0}^{10}\binom{10-k}{3}=\sum_{i=0}^{10}\binom{i}{3}=\binom{11}{4}=330.$$
Thus
$$P(3\,black)=\frac{1}{11}\times\frac{330}{120}=\frac{330}{1320}=\frac{1}{4}.$$
Step 5: Apply Bayes theorem:
$$P(k=1\mid3\,black)=\frac{P(k=1,3\,black)}{P(3\,black)}=\frac{7/110}{1/4}=\frac{7}{110}\times4=\frac{28}{110}=\frac{14}{55}.$$

Use the identity $$\sum_{i=0}^{n}\binom{i}{r}=\binom{n+1}{r+1}$$ to compute the denominator quickly.

- Forgetting the uniform prior over k
- Miscomputing the sum of combinations in the denominator

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