Question

There are ten urns of which each of three contain 1 white and 8 black balls; each of other three contains 9 white and 1 black balls, and of the remaining four each contains 5 white and 5 black balls. One of the urns is selected at random and a ball taken blindly from it and it turns outto be white. What is the probability that an urn containing 1 white and 9 black balls was selected?​

Answer 1

Given:

There are 10 urns. Each urn contains 10 balls.

To find:

The probability that an urn containing 1 white and 9 black ball was selected.

Solution:

Let U be the Urn.

Let E be the event of the ball picked.

Let U1 = {1 white ball, 8 black ball}

Let U2 = {9 white ball, 1 black ball}

Let U3 ={5white ball, 5 black ball}

Therefore, P(U1)=P(U2)=P(U3)= 1/10

Let E be the event that a ball is chosen from urn U, E1 be the event that a ball is chosen from urn U1 and E2 be the event that a ball is chosen from urn U2 and E3 be the event that a ball is chosen from urn U3.

So, P(E1)=P(E2)=P(E3)=1/10

Let E be the event of white ball drawn

[tex]P(\frac{E}{E1})=\frac{1}{10}\\ P(\frac{E}{E2})=\frac{9}{10}\\ P(\frac{E}{E3})=\frac{5}{10} [/tex]

$Let, \[P(E/E1)=\frac{P(E1)\times P(E/E1)}{P(E1)\times P(E/E1)+P(E2)\times P(E/E2)+P(E3)\times P(E/E3)}\]$

                 

Here, we will substitute the terms and find the value

$\[=\frac{\frac{1}{10}\times \frac{9}{10}}{\frac{1}{10}\times \frac{1}{10}+\frac{1}{10}\times \frac{9}{10}+\frac{1}{10}\times \frac{5}{10}}\]$

[tex]\[=\frac{\frac{9}{100}}{\frac{1}{100}+\frac{9}{100}+\frac{5}{100}}\] \[=\frac{9}{15}\] \[=\frac{3}{5}\] [/tex]

Final Answer:

The probability that an urn containing 1 white and 9 black ball was selected is  $\frac{3}{5}$

[tex]\[\begin{align} & =\frac{\frac{1}{10}\times \frac{9}{10}}{\frac{1}{10}\times \frac{1}{10}+\frac{1}{10}\times \frac{9}{10}+\frac{1}{10}\times \frac{5}{10}} \\ & =\frac{\frac{9}{100}}{\frac{1}{100}+\frac{9}{100}+\frac{5}{100}} \\ & =\frac{9}{15} \\ & =\frac{3}{5} \\ \end{align}\] [/tex][tex]\[\begin{align} & =\frac{\frac{1}{10}\times \frac{9}{10}}{\frac{1}{10}\times \frac{1}{10}+\frac{1}{10}\times \frac{9}{10}+\frac{1}{10}\times \frac{5}{10}} \\ & =\frac{\frac{9}{100}}{\frac{1}{100}+\frac{9}{100}+\frac{5}{100}} \\ & =\frac{9}{15} \\ & =\frac{3}{5} \\ \end{align}\] [/tex]