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  • There are three urns containing 2 white and 3 black balls, 3 white and 2 black balls, and 4 white and 1 black balls, respectively. There is an equal probability of each urn being chosen. A ball is drawn at random from the chosen urn and it is found to be

There are three urns containing 2 white and 3 black balls, 3 white and 2 black balls, and 4 white and 1 black balls, respectively. There is an equal probability of each urn being chosen. A ball is drawn at random from the chosen urn and it is found to be white. Find the probability that the ball drawn was from the second urn.

Answers (1)

Solution

Given-

There are 3 urns U1, U2 and U3

Let U1 be 2 white and 3 black balls

U2 be 3 white and 2 black balls
U3 be 4 white and 1 black balls
Therefore, Total balls = 5

As there is an equal probability of each urn being chosen

\therefore \mathrm{P}\left(\mathrm{U}_{1}\right)=\mathrm{P}\left(\mathrm{U}_{2}\right)=\mathrm{P}\left(\mathrm{U}_{3}\right)=\frac{1}{3}
Let E_{1}, E_{2}$ and $E_{3} be the event that a ball is chosen from an urn U_{1}, \\\mathrm{U}_{2}$ and $\mathrm{U}_{3}  respectively.
\\\therefore P\left(E_{1}\right)=P\left(E_{2}\right)=P\left(E_{3}\right)=\frac{1}{3}

Let A be the event that the white ball is drawn.
P(A|E1) is the probability that the white ball is chosen from urn U1
P(A|E2) is the probability that the white ball is chosen from urn U2
P(A|E3) is the probability that the white ball is chosen from urn U3

\therefore P\left(A \mid E_{1}\right)=\frac{2}{5}, P\left(A \mid E_{2}\right)=\frac{3}{5} and P\left(A \mid E_{3}\right)=\frac{4}{5}

To find- the probability that the ball is drawn was from \mathrm{U}_{2}.
Using Bayes' theorem to find the probability of occurrence of an event A when event B has already occurred.
\underset{\therefore}{\mathbf{P}}(\mathbf{A} \mid \mathbf{B})=\frac{P(A) P(B \mid A)}{P(B)}

\\ \mathrm{P}\left(\mathrm{E}_{2} \mid \mathrm{A}\right)$ is the probability that white ball is selected from urn $\mathrm{U}_{2}

\\ P\left(E_{2} \mid A\right)=\frac{P\left(E_{2}\right) \times P\left(A \mid E_{2}\right)}{P\left(E_{1}\right) \times P\left(A \mid E_{1}\right)+P\left(E_{2}\right) \times P\left(A \mid E_{2}\right)+P\left(E_{3}\right) \times P\left(A \mid E_{3}\right)} \\ =\frac{\frac{1}{3} \times \frac{3}{5}}{\frac{1}{3} \times \frac{2}{5}+\frac{1}{3} \times \frac{3}{5}+\frac{1}{3} \times \frac{4}{5}} \\ =\frac{\frac{1}{5}}{\frac{2}{15}+\frac{3}{15}+\frac{4}{15}} \\ =\frac{\frac{1}{5}}{\frac{9}{15}} \\ =\frac{1}{3}

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