Question

A bag contains some white and some black balls, all combinations of balls being equally likely. The total number of balls in the bag is 10. If three balls are drawn at random without replacement and all of them are found to be black, what is the probability that the bag contains 1 white and 9 black balls?
(a). \[\dfrac{{14}}{{55}}\]
(b). \[\dfrac{{12}}{{55}}\]
(c). \[\dfrac{2}{{11}}\]
(d). \[\dfrac{8}{{55}}\]

Answer 1

Hint: Assign variables to the events for all combinations of balls in the bag. Then find the probability that 3 black balls are drawn using the total probability rule. Then find the probability that the bag contains 1 white and 9 black balls using Bayes’ theorem.

Complete step-by-step answer:
It is given that the bag contains some white balls and some black balls, to a total of 10 balls, with all combinations being equally likely.
Let the event \[{E_i}\] denote the event that the bag contains \[i\] black balls and \[10 - i\] white balls, where \[i\] can take the values from 0 to 10. Then we have:
\[P({E_i}) = \dfrac{1}{{11}}.............(1)\]
Let A be the event that 3 black balls are drawn. Then, we have:
\[P(A|{E_0}) = P(A|{E_1}) = P(A|{E_2}) = 0\]
\[P(A|{E_i}) = \dfrac{{{}^i{C_3}}}{{{}^{10}{C_3}}}{\text{ (}}i > 2{\text{)}}.............{\text{(2)}}\]
By the theorem of total probability, the total probability of drawing 3 black balls at random is given by:
\[P(A) = \sum\limits_{i = 0}^{10} {P({E_i})P(A|{E_i})} \]
Substituting equation (1) and equation (2), we have:
\[P(A) = \dfrac{1}{{11}}.\dfrac{1}{{{}^{10}{C_3}}}({}^3{C_3} + {}^4{C_3} + .......... + {}^{10}{C_3})\]
We can write \[{}^3{C_3}\] as \[{}^4{C_4}\], then we have:
\[P(A) = \dfrac{1}{{11}}.\dfrac{1}{{{}^{10}{C_3}}}({}^4{C_4} + {}^4{C_3} + .......... + {}^{10}{C_3})\]
We have the following formula:
\[{}^n{C_r} + {}^n{C_{r + 1}} = {}^{n + 1}{C_{r + 1}}............(3)\]
Using the formula (3), we have:
\[P(A) = \dfrac{1}{{11}}.\dfrac{1}{{{}^{10}{C_3}}}(({}^4{C_4} + {}^4{C_3}) + .......... + {}^{10}{C_3})\]
\[P(A) = \dfrac{1}{{11}}.\dfrac{1}{{{}^{10}{C_3}}}({}^5{C_4} + {}^5{C_3} + .......... + {}^{10}{C_3})\]
Similarly, using recursion, we have:
\[P(A) = \dfrac{1}{{11}}.\dfrac{1}{{{}^{10}{C_3}}}.{}^{11}{C_4}\]
Evaluating the expression, we have:
\[P(A) = \dfrac{1}{{11}}.\dfrac{{3!7!}}{{10!}}.\dfrac{{11!}}{{4!7!}}\]
\[P(A) = \dfrac{1}{4}............(4)\]
\[P({E_9}|A)\] is the probability that the bag contains 1 white and 9 black balls given that 3 balls drawn at random are black. Then, by Bayes’ theorem, we have:
\[P({E_9}|A) = \dfrac{{P({E_9})P(A|{E_9})}}{{P(A)}}\]
Using equations (1), (2), and (4), we have:
\[P({E_9}|A) = \dfrac{{\dfrac{1}{{11}}\dfrac{{{}^9{C_3}}}{{{}^{10}{C_3}}}}}{{\dfrac{1}{4}}}\]
Simplifying, we have:
\[P({E_9}|A) = \dfrac{{\dfrac{1}{{11}}\dfrac{{9!}}{{6!3!}}\dfrac{{7!3!}}{{10!}}}}{{\dfrac{1}{4}}}\]
\[P({E_9}|A) = \dfrac{{\dfrac{1}{{11}}.\dfrac{7}{{10}}}}{{\dfrac{1}{4}}}\]
\[P({E_9}|A) = \dfrac{{\dfrac{1}{{11}}.\dfrac{7}{5}}}{{\dfrac{1}{2}}}\]
\[P({E_9}|A) = \dfrac{{14}}{{55}}\]
Hence, the correct answer is option (a).

Note: It is tricky to evaluate the total probability using the properties of combinations. If you directly know the formula, then you can substitute and solve it directly.