Question

There are 3 boxes, the first one containing 1 white, 2 red and 3 black balls; the second one containing 2 white, 3 red and 1 black ball and the third one containing 3 white, 1 red and 2 black balls. A box is chosen at random and from it two balls are drawn at random. One ball is red and the other, white. What is the probability that they come from the second box?

Answer 1

Here we will find the probability by using Bayes’ Theorem. For that, we will first find the probability that the balls drawn are one white and one red and it is from box 1. Similarly, we will first find the probability that the balls drawn are one white and one red and it is from box 2 and then we will first find the probability that the balls drawn are one white and one red and it is from box 3. Then we will use the formula of Bayes’ Theorem to find the probability.

Complete step by step solution:
Let E1, E2 and E3 be the three events that the balls are drawn from box1, box 2 and box 3 respectively, and let E be the event that the balls drawn are one white and one red.
Therefore,
$P\left( {{E}_{1}} \right)=P\left( {{E}_{1}} \right)=P\left( {{E}_{1}} \right)=\dfrac{1}{3}$
Now, we will find the probability that the balls drawn are one white and one red and it is from box 1
$\Rightarrow$ $P\left( {}^{E}/{}_{{{E}_{1}}} \right)=\dfrac{{}^{1}{{C}_{1}}\times {}^{2}{{C}_{1}}}{{}^{6}{{C}_{2}}}$
On further simplification, we get
$\Rightarrow$ $P\left( {}^{E}/{}_{{{E}_{1}}} \right)=\dfrac{1\times 2}{15}=\dfrac{2}{15}$
Again, we will find the probability that the balls drawn are one white and one red and it is from box 2
$\Rightarrow$ $P\left( {}^{E}/{}_{{{E}_{2}}} \right)=\dfrac{{}^{2}{{C}_{1}}\times {}^{3}{{C}_{1}}}{{}^{6}{{C}_{2}}}$
On further simplification, we get
$\Rightarrow$ $P\left( {}^{E}/{}_{{{E}_{2}}} \right)=\dfrac{2\times 3}{15}=\dfrac{6}{15}$
Again, we will find the probability that the balls drawn are one white and one red and it is from box 3
$\Rightarrow$ $P\left( {}^{E}/{}_{{{E}_{3}}} \right)=\dfrac{{}^{3}{{C}_{1}}\times {}^{1}{{C}_{1}}}{{}^{6}{{C}_{2}}}$
On further simplification, we get
$\Rightarrow$ $P\left( {}^{E}/{}_{{{E}_{3}}} \right)=\dfrac{3\times 1}{15}=\dfrac{3}{15}=\dfrac{1}{5}$
We will use Bayes’ Theorem now.
According to the theorem, the probability that the ball drawn are from box 2 and the ball drawn are one white and one red is represented by $P\left( {}^{{{E}_{1}}}/{}_{E} \right)$ and it will be equal to
$\Rightarrow$ $\dfrac{P\left( {}^{E}/{}_{{{E}_{1}}} \right).P\left( {{E}_{1}} \right)}{P\left( {}^{E}/{}_{{{E}_{1}}} \right).P\left( {{E}_{1}} \right)+P\left( {}^{E}/{}_{{{E}_{2}}} \right).P\left( {{E}_{2}} \right)+P\left( {}^{E}/{}_{{{E}_{3}}} \right).P\left( {{E}_{3}} \right)}$
We will put the values that we have calculated in the formula.
$\Rightarrow$ $P\left( {}^{{{E}_{1}}}/{}_{E} \right)=\dfrac{\dfrac{2}{15}.\dfrac{1}{3}}{\dfrac{2}{15}.\dfrac{1}{3}+\dfrac{6}{15}.\dfrac{1}{3}+\dfrac{1}{5}.\dfrac{1}{3}}$
On multiplying the terms in numerator and denominator, we get
$\Rightarrow$ $P\left( {}^{{{E}_{1}}}/{}_{E} \right)=\dfrac{\dfrac{2}{45}}{\dfrac{2}{45}+\dfrac{6}{45}+\dfrac{1}{15}}$
Simplifying the fraction, we get
$\Rightarrow$ $P\left( {}^{{{E}_{1}}}/{}_{E} \right)=\dfrac{2}{11}$
Thus, the required probability is $\dfrac{2}{11}$.

Note: We have calculated the given probability by Bayes’ Theorem. Bayes’ Theorem is a mathematical formula to calculate or to determine 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. Bayes' theorem is used to determine the accuracy of medical test results.