Question

A bag contains some white balls and some black balls, all combinations being equally likely. The total number of balls in the bag is 12. Four balls are drawn at random without replacement.
Find the following:
(a) Probability that all the balls are black is equal to
(b) If the bag contains 10 black and 2 white balls, then the probability that all four balls are black is equal to
(c) If all the four balls are black, then the probability that the bag contains 10 black balls is equal to

Answer 1

In this question, we can find the probability of drawing 4 black balls from 10 black and 2 white balls using the combination formula ${}^{n}{{C}_{r}}$ and the probability formula $\dfrac{m}{n}$. Then we can find the probability that all balls are black using the total probability formula given by $P\left( A \right)=\sum\limits_{r=1}^{n}{P\left( {{E}_{r}} \right)P\left( \dfrac{A}{{{E}_{r}}} \right)}$.Now, we need to find the probability that after drawing 4 black balls it should contain 10 black balls which can be done using conditional probability formula $P\left( \dfrac{A}{B} \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)}$

Complete step-by-step solution -
Now, let us find the probability of drawing 4 black balls from 10 black balls and 2 white balls
Number of favourable outcomes in this case will be the number of combinations possible for 4 back balls out of 10
Total number of outcomes will be the number of combinations of 4 balls out of 12 balls
$\Rightarrow P=\dfrac{m}{n}$
As we already know that formula for combinations is ${}^{n}{{C}_{r}}$ an don substituting the respective values we get,
$\Rightarrow P=\dfrac{{}^{10}{{C}_{4}}}{{}^{12}{{C}_{4}}}$
Now, this can be also written as
$\Rightarrow P=\dfrac{10!}{6!4!}\times \dfrac{8!4!}{12!}$
Now, on further simplification we get,

& \Rightarrow P=\dfrac{8\times 7}{12\times 11} \\\ & \therefore P=\dfrac{14}{33} \\\ \end{aligned}$$ Let Ei be an event in which there are i black balls then we have 12-i white balls, where $$i=0,1,2,3,...12$$i.e., we have 13 outcomes possible with each combination equally likely Let us assume the event that all the four balls are black as A As we already know the formula of total probability is given by $$P\left( A \right)=\sum\limits_{r=1}^{n}{P\left( {{E}_{r}} \right)P\left( \dfrac{A}{{{E}_{r}}} \right)}$$ Now, using the above formula we can find the probability for all the balls to be black $$P\left( {{E}_{i}} \right)=\dfrac{1}{13}$$ Now, when $$i=0,1,2,3$$then $$P\left( \dfrac{A}{{{E}_{i}}} \right)=0$$ because we need to draw 4 black balls form the bag Now, when $$i=4,5,....,12$$ having total number of outcomes as 9 we have $$P\left( \dfrac{A}{{{E}_{i}}} \right)=\dfrac{{}^{i}{{C}_{4}}}{{}^{9}{{C}_{4}}}$$ Now, from the total probability formula we get, $$\Rightarrow P\left( A \right)=\sum\limits_{i=4}^{12}{P\left( {{E}_{i}} \right)P\left( \dfrac{A}{{{E}_{i}}} \right)}$$ Now, on substituting the respective values and expanding it we get, $$\Rightarrow P\left( A \right)=\dfrac{1}{13}\left( \dfrac{{}^{4}{{C}_{4}}}{{}^{9}{{C}_{4}}}+\dfrac{{}^{5}{{C}_{4}}}{{}^{9}{{C}_{4}}}+\dfrac{{}^{6}{{C}_{4}}}{{}^{9}{{C}_{4}}}+\dfrac{{}^{7}{{C}_{4}}}{{}^{9}{{C}_{4}}}+\dfrac{{}^{8}{{C}_{4}}}{{}^{9}{{C}_{4}}}+\dfrac{{}^{9}{{C}_{4}}}{{}^{9}{{C}_{4}}} \right)$$ Here, we can write $${}^{4}{{C}_{4}}$$ as $${}^{5}{{C}_{5}}$$ because both of them have the equal value which is one so we can replace $${}^{4}{{C}_{4}}$$ with $${}^{5}{{C}_{5}}$$ to simplify it further As we already know that $${}^{n}{{C}_{r}}+{}^{n}{{C}_{r-1}}={}^{n+1}{{C}_{r}}$$ Now, using this formula the above expression can be further written as $$\Rightarrow P\left( A \right)=\dfrac{1}{13}\left( \dfrac{{}^{5}{{C}_{5}}}{{}^{9}{{C}_{4}}}+\dfrac{{}^{5}{{C}_{4}}}{{}^{9}{{C}_{4}}}+\dfrac{{}^{6}{{C}_{4}}}{{}^{9}{{C}_{4}}}+\dfrac{{}^{7}{{C}_{4}}}{{}^{9}{{C}_{4}}}+\dfrac{{}^{8}{{C}_{4}}}{{}^{9}{{C}_{4}}}+\dfrac{{}^{9}{{C}_{4}}}{{}^{9}{{C}_{4}}}+\dfrac{{}^{10}{{C}_{4}}}{{}^{9}{{C}_{4}}}+\dfrac{{}^{11}{{C}_{4}}}{{}^{9}{{C}_{4}}}+\dfrac{{}^{12}{{C}_{4}}}{{}^{9}{{C}_{4}}} \right)$$ Now, this can be further written as $$\Rightarrow P\left( A \right)=\dfrac{1}{13}\left( \dfrac{{}^{6}{{C}_{5}}}{{}^{9}{{C}_{4}}}+\dfrac{{}^{6}{{C}_{4}}}{{}^{9}{{C}_{4}}}+\dfrac{{}^{7}{{C}_{4}}}{{}^{9}{{C}_{4}}}+\dfrac{{}^{8}{{C}_{4}}}{{}^{9}{{C}_{4}}}+\dfrac{{}^{9}{{C}_{4}}}{{}^{9}{{C}_{4}}}+\dfrac{{}^{10}{{C}_{4}}}{{}^{9}{{C}_{4}}}+\dfrac{{}^{11}{{C}_{4}}}{{}^{9}{{C}_{4}}}+\dfrac{{}^{12}{{C}_{4}}}{{}^{9}{{C}_{4}}} \right)$$ Now, on further simplification we get, $$\Rightarrow P\left( A \right)=\dfrac{1}{13}\left( \dfrac{{}^{7}{{C}_{5}}}{{}^{9}{{C}_{4}}}+\dfrac{{}^{7}{{C}_{4}}}{{}^{9}{{C}_{4}}}+\dfrac{{}^{8}{{C}_{4}}}{{}^{9}{{C}_{4}}}+\dfrac{{}^{9}{{C}_{4}}}{{}^{9}{{C}_{4}}}+\dfrac{{}^{10}{{C}_{4}}}{{}^{9}{{C}_{4}}}+\dfrac{{}^{11}{{C}_{4}}}{{}^{9}{{C}_{4}}}+\dfrac{{}^{12}{{C}_{4}}}{{}^{9}{{C}_{4}}} \right)$$ Similarly, using the same formula we can further write it as $$\Rightarrow P\left( A \right)=\dfrac{1}{13}\left( \dfrac{{}^{8}{{C}_{5}}}{{}^{9}{{C}_{4}}}+\dfrac{{}^{8}{{C}_{4}}}{{}^{9}{{C}_{4}}}+\dfrac{{}^{9}{{C}_{4}}}{{}^{9}{{C}_{4}}}+\dfrac{{}^{10}{{C}_{4}}}{{}^{9}{{C}_{4}}}+\dfrac{{}^{11}{{C}_{4}}}{{}^{9}{{C}_{4}}}+\dfrac{{}^{12}{{C}_{4}}}{{}^{9}{{C}_{4}}} \right)$$ Again, on further simplification similarly we get, $$\Rightarrow P\left( A \right)=\dfrac{1}{13}\left( \dfrac{{}^{13}{{C}_{5}}}{{}^{9}{{C}_{4}}} \right)$$ Now, on further simplification we get, $$\Rightarrow P\left( A \right)=\dfrac{1}{13}\times \dfrac{13!}{8!5!}\times \dfrac{5!4!}{9!}$$ Now, this can also be written as $$\Rightarrow P\left( A \right)=\dfrac{1}{13}\times \dfrac{13\times 12\times 11\times 10}{8\times 7\times 6\times 5}$$ $$\therefore P\left( A \right)=\dfrac{11}{14}$$ Now, we have to find the probability of the bag to contain 10 back balls when there are 4 black balls already drawn As we already know the formula for conditional probability that $$P\left( \dfrac{A}{B} \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)}$$ Here, we need to find the value of $$P\left( \dfrac{{{E}_{10}}}{A} \right)$$ $$\Rightarrow P\left( \dfrac{{{E}_{10}}}{A} \right)=\dfrac{P\left( {{E}_{10}}\cap A \right)}{P\left( A \right)}$$ Now, this can also be written as $$\Rightarrow P\left( \dfrac{{{E}_{10}}}{A} \right)=\dfrac{P\left( {{E}_{10}} \right)\times P\left( \dfrac{A}{{{E}_{10}}} \right)}{P\left( A \right)}$$ Now, let us substitute the respective values in the above expression $$\Rightarrow P\left( \dfrac{{{E}_{10}}}{A} \right)=\dfrac{\dfrac{1}{13}\times \dfrac{{}^{10}{{C}_{4}}}{{}^{9}{{C}_{4}}}}{\dfrac{11}{14}}$$ Now, on further simplification we get, $$\Rightarrow P\left( \dfrac{{{E}_{10}}}{A} \right)=\dfrac{1}{13}\times \dfrac{10!}{6!4!}\times \dfrac{5!4!}{9!}\times \dfrac{14}{11}$$ Now, on cancelling out the common terms and simplifying further we get, $$\begin{aligned} & \Rightarrow P\left( \dfrac{{{E}_{10}}}{A} \right)=\dfrac{14\times 10}{13\times 11\times 6} \\\ & \therefore P\left( \dfrac{{{E}_{10}}}{A} \right)=\dfrac{70}{429} \\\ \end{aligned}$$ **Note:** It is important to note that while finding the probability of all the balls drawn to be black we need to keep in mind that when $$i=0,1,2,3$$ then the combinations cannot be found as i is of lesser value. So, the respective probability will be zero. It is also to be noted that while finding the probability for the bag to contain 10 black balls we need to note that this event occurs only after the happening of event A. Hence, we cannot directly find its value so we use the conditional probability as a certain condition is given.