Question

There are two boxes $I$ and $II$ . Box $I$ contains 3 red and 6 black balls. Box $I$ contains 5 red and $'n'$ black balls. One of the two boxes, box $I$ and box $II$ is selected at random and a ball is drawn at random. The ball drawn is found to be red. If the probability that this red ball comes out from the box $II$ is $\dfrac{3}{5}$ ​, find the value of $'n'$ .

Answer 1

Here we need to find the number of black balls present in the second box. We will use the Baye’s theorem of the probability. We will first find the probability that the balls drawn are red and it is from box 1. Similarly, we will find the probability that the balls drawn is red and it is from box 2. Then substituting all the values in the formula of Baye’s theorem we will get the value of the required variable. Baye’s theorem is also known as the conditional probability.

Formula Used:
We will use the Baye’s theorem which is given by the formula, $P\left( {\dfrac{{{E_2}}}{E}} \right) = \dfrac{{P\left( {\dfrac{E}{{{E_2}}}} \right).P\left( {{E_2}} \right)}}{{P\left( {\dfrac{E}{{{E_1}}}} \right).P\left( {{E_1}} \right) + P\left( {\dfrac{E}{{{E_2}}}} \right).P\left( {{E_2}} \right)}}$.

Complete step-by-step answer:
Let ${E_2}$ and${E_1}$ be the two events that the ball are drawn from box1 and box 2 respectively, and let $E$ be the event that he balls drawn is red.
Therefore,
$P\left( {{E_1}} \right) = P\left( {{E_2}} \right) = \dfrac{1}{2}$
Now, we will find the probability that the balls drawn is red and it is from box 1
$P\left( {\dfrac{E}{{{E_1}}}} \right) = \dfrac{{{}^3{C_1}}}{{{}^9{C_1}}}$
Now, simplifying the above expression , using the formula ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, we get
$\Rightarrow P\left( {\dfrac{E}{{{E_1}}}} \right) = \dfrac{{\dfrac{{3!}}{{1!\left( {3 - 1} \right)!}}}}{{\dfrac{{9!}}{{1!\left( {9 - 1} \right)!}}}}$
Computing the factorials of the numbers, we get
$\Rightarrow P\left( {\dfrac{E}{{{E_1}}}} \right) = \dfrac{{\dfrac{{3 \times 2 \times 1}}{{1 \times 2 \times 1}}}}{{\dfrac{{9 \times 8!}}{{1 \times 8!}}}}$
On further simplification, we get
$\Rightarrow P\left( {\dfrac{E}{{{E_1}}}} \right) = \dfrac{1}{3}$
Now, we will find the probability that the balls drawn is red and it is from box 1
$P\left( {\dfrac{E}{{{E_2}}}} \right) = \dfrac{{{}^5{C_1}}}{{{}^{5 + n}{C_1}}}$
Now, simplifying the above expression , using the formula ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, we get
$\Rightarrow P\left( {\dfrac{E}{{{E_2}}}} \right) = \dfrac{{\dfrac{{5!}}{{1!\left( {5 - 1} \right)!}}}}{{\dfrac{{\left( {5 + n} \right)!}}{{1!\left( {5 + n - 1} \right)!}}}}$
Computing the factorials of the numbers, we get
$\Rightarrow P\left( {\dfrac{E}{{{E_2}}}} \right) = \dfrac{{\dfrac{{5 \times 4!}}{{1!4!}}}}{{\dfrac{{\left( {5 + n} \right) \times \left( {4 + n} \right)!}}{{\left( {4 + n} \right)!}}}}$
On further simplification, we get
$\Rightarrow P\left( {\dfrac{E}{{{E_2}}}} \right) = \dfrac{5}{{5 + n}}$
We will use Bayes’ Theorem now.
According to the theorem, the probability that the ball drawn are from box 2 and the ball drawn is red is represented by $P\left( {\dfrac{{{E_2}}}{E}} \right)$ and it will be equal to $\dfrac{{P\left( {\dfrac{E}{{{E_2}}}} \right).P\left( {{E_2}} \right)}}{{P\left( {\dfrac{E}{{{E_1}}}} \right).P\left( {{E_1}} \right) + P\left( {\dfrac{E}{{{E_2}}}} \right).P\left( {{E_2}} \right)}}$.
Substituting all the values of the probability in the formula $P\left( {\dfrac{{{E_2}}}{E}} \right) = \dfrac{{P\left( {\dfrac{E}{{{E_2}}}} \right).P\left( {{E_2}} \right)}}{{P\left( {\dfrac{E}{{{E_1}}}} \right).P\left( {{E_1}} \right) + P\left( {\dfrac{E}{{{E_2}}}} \right).P\left( {{E_2}} \right)}}$, we get
$P\left( {\dfrac{{{E_2}}}{E}} \right) = \dfrac{{\dfrac{1}{2}.\dfrac{5}{{5 + n}}}}{{\dfrac{1}{2}.\dfrac{1}{3} + \dfrac{1}{2}.\dfrac{5}{{5 + n}}}}$ ………. $\left( 1 \right)$
It is given that the probability that this red ball comes out from box $II$ is $\dfrac{3}{5}$.
Substituting $P\left( {\dfrac{{{E_2}}}{E}} \right) = \dfrac{3}{5}$ in the above equation, we get
$\Rightarrow \dfrac{3}{5} = \dfrac{{\dfrac{1}{2}.\dfrac{5}{{5 + n}}}}{{\dfrac{1}{2}.\dfrac{1}{3} + \dfrac{1}{2}.\dfrac{5}{{5 + n}}}}$
On further simplification, we get
$\Rightarrow \dfrac{3}{5} = \dfrac{{\dfrac{5}{{5 + n}}}}{{\dfrac{1}{3} + \dfrac{5}{{5 + n}}}}$
On cross multiplying the terms, we get
$\Rightarrow \dfrac{3}{5}\left( {\dfrac{1}{3} + \dfrac{5}{{5 + n}}} \right) = \dfrac{5}{{5 + n}}$
On multiplying the terms, we get
$\Rightarrow \dfrac{1}{5} + \dfrac{3}{{5 + n}} = \dfrac{5}{{5 + n}}$
Now, we will subtract the term $\dfrac{5}{{5 + n}}$ from sides. So,
$\begin{array}{l} \Rightarrow \dfrac{1}{5} = \dfrac{5}{{5 + n}} - \dfrac{3}{{5 + n}}\\\ \Rightarrow \dfrac{1}{5} = \dfrac{{5 - 3}}{{5 + n}}\\\ \Rightarrow \dfrac{1}{5} = \dfrac{2}{{5 + n}}\end{array}$
On cross multiplying the terms, we get
$\Rightarrow 5 + n = 10$
Now, we will subtract the number 5 from both sides.
$\begin{array}{l} \Rightarrow 5 + n - 5 = 10 - 5\\\ \Rightarrow n = 5\end{array}$

Note: Bayes’ Theorem is defined as a mathematical formula to determine or to calculate the conditional probability. Bayes' theorem is also known as Bayes' Rule or Bayes' Law and it is the foundation of the field of Bayesian statistics. Probability is defined as the certainty of occurrence of an event. Probability of an event always lies between 0 to 1. If the probability is 1 then it is known as a sure event and the event having a probability 0 can never occur.