Question

The contents of urn I and II are as follows,
Urn I: 4 white and 5 black balls
Urn II: 3 white and 6 black balls
One urn is chosen at random and a ball is drawn, and its colour is noted and replaced back to the urn. Again, a ball is drawn from the same urn, colour is noted and replaced. The process is repeated 4 times and as a result one ball of white colour and 3 of black colour are noted. The probability that the chosen urn was I is
(a) $\dfrac{125}{287}$
(b) $\dfrac{64}{127}$
(c) $\dfrac{25}{287}$
(d) \[\dfrac{79}{192}\]
(e) \[\dfrac{250}{{{9}^{4}}}\]

Answer 1

Hint: First, we will assume that let $\text{P}\left( \text{A} \right)$ i.e. probability that one ball drawn is white and three balls are black from urn I. So, this is given by the formula $\text{P}\left( \text{A} \right)\text{=P}\left( \text{I} \right)\text{ }\!\!\times\!\!\text{ P}\left( \dfrac{\text{A}}{\text{I}} \right)\text{+P}\left( \text{II} \right)\text{ }\!\!\times\!\!\text{ P}\left( \dfrac{\text{A}}{\text{II}} \right)$ . After finding this value, we have to find probability of $\text{P}\left( \dfrac{\text{I}}{\text{A}} \right)$ i.e. balls selected are from urn I out of all the total probability of selection. This can be obtained by using the formula $\text{P}\left( \dfrac{\text{I}}{\text{A}} \right)\text{=}\dfrac{\text{P}\left( \text{I} \right)\text{ }\!\!\times\!\!\text{ P}\left( \dfrac{\text{A}}{\text{I}} \right)}{\text{P}\left( \text{A} \right)}$ . Thus, the required answer will be obtained.

Complete step by step solution:
Here, we will assume that let I be the event that urn I is selected and II be the event that urn II is selected. Now, we will take the let A be the event that one ball drawn is white and three balls are black from urn I. So, we have to find probability as $\text{P}\left( \dfrac{\text{I}}{\text{A}} \right)$ .
First, we will find $\text{P}\left( \text{A} \right)$ i.e. probability that one ball drawn is white and three balls are black from urn I. So, this is given by the formula
$\text{P}\left( \text{A} \right)\text{=P}\left( \text{I} \right)\text{ }\!\!\times\!\!\text{ P}\left( \dfrac{\text{A}}{\text{I}} \right)\text{+P}\left( \text{II} \right)\text{ }\!\!\times\!\!\text{ P}\left( \dfrac{\text{A}}{\text{II}} \right)$ ………………………(1)
So, here there are two urn I and II, so there individual probability will be equal to 0.5 i.e. $\dfrac{1}{2}$ as we know that maximum probability is equal to 1.
 On substituting the values in equation (1), we get
$\text{P}\left( \text{A} \right)\text{=P}\left( \text{I} \right)\text{ }\!\!\times\!\!\text{ P}\left( \dfrac{\text{A}}{\text{I}} \right)\text{+P}\left( \text{II} \right)\text{ }\!\!\times\!\!\text{ P}\left( \dfrac{\text{A}}{\text{II}} \right)$
Here, we have to find the probability of one white ball i.e. in urn I there are 4 white balls given. So, for that probability will $\dfrac{4}{9}$ i.e. $\text{P}\left( \dfrac{\text{A}}{\text{I}} \right)$ and three black balls to be selected so, it is written as ${{\left( \dfrac{5}{9} \right)}^{3}}$because 3 black balls are to be selected like after selecting one black ball it is to be kept inside the urn and then again selecting balls. Similarly, we can write $\text{P}\left( \dfrac{\text{A}}{\text{II}} \right)=\dfrac{3}{9}$ and there will be ${{\left( \dfrac{6}{9} \right)}^{3}}$ for selecting black balls. This whole selection process is repeated 4 times so multiplying by 4. In the denominator, there will be summation of both black and white ball i.e. $4+5=9$ .
Thus, we get as
$\text{P}\left( \text{A} \right)\text{=}\left( \dfrac{1}{2}\cdot \dfrac{4}{9}{{\left( \dfrac{5}{9} \right)}^{3}}\times \text{4} \right)\text{+}\left( \dfrac{1}{2}\cdot \dfrac{3}{9}{{\left( \dfrac{6}{9} \right)}^{3}}\text{ }\!\!\times\!\!\text{ 4} \right)$
So, on further solving the equation, we get as
$\text{P}\left( \text{A} \right)\text{=}\dfrac{\text{125}\times \text{4}}{2\times {{9}^{4}}}\times \text{4+}\dfrac{\text{216}\times \text{3}}{2\times {{9}^{4}}}\times 4$
$\text{P}\left( \text{A} \right)\text{=}\dfrac{500}{2\times {{9}^{4}}}\times \text{4+}\dfrac{648}{2\times {{9}^{4}}}\times 4$
Now taking denominator and 4 in multiplication common so, we will get
$\text{P}\left( \text{A} \right)\text{=}\dfrac{4}{2\times {{9}^{4}}}\left( 500+648 \right)=\dfrac{1148}{2\times {{9}^{4}}}\times 4$ ……………………………….(2)
Now, we have to find the probability of one white and three black balls from urn I denoted as $\text{P}\left( \dfrac{\text{I}}{\text{A}} \right)$ .
Therefore, we can write this as
$\text{P}\left( \dfrac{\text{I}}{\text{A}} \right)\text{=}\dfrac{\text{P}\left( \text{I} \right)\text{ }\!\!\times\!\!\text{ P}\left( \dfrac{\text{A}}{\text{I}} \right)}{\text{P}\left( \text{A} \right)}$
Now, we will put the values as already found in the above equation. So, we get as
$\text{P}\left( \dfrac{\text{I}}{\text{A}} \right)\text{=}\dfrac{\left( \dfrac{1}{2}\cdot \dfrac{4}{9}{{\left( \dfrac{5}{9} \right)}^{3}}\times 4 \right)}{\dfrac{1148}{2\times {{9}^{4}}}\times 4}$
On further simplification, we get
$\text{P}\left( \dfrac{\text{I}}{\text{A}} \right)\text{=}\dfrac{\left( \dfrac{500}{2\times {{9}^{4}}}\times 4 \right)}{\dfrac{1148}{2\times {{9}^{4}}}\times 4}$
Now, we will be cancelling the denominator part and 4 which is in multiplication, we get
$\text{P}\left( \dfrac{\text{I}}{\text{A}} \right)\text{=}\dfrac{500}{1148}=\dfrac{125}{287}$
Thus, option (a) is the correct answer.

Note: Be careful while applying the notation about what is asked in question. Here, we have to find probability $\text{P}\left( \dfrac{\text{I}}{\text{A}} \right)$ i.e. selection should be from urn I out of total selection but by mistake if written as $\text{P}\left( \dfrac{\text{A}}{\text{I}} \right)$ then whole meaning will be changed i.e. $\text{P}\left( \dfrac{\text{A}}{\text{I}} \right)\text{=}\dfrac{\text{P}\left( \text{A} \right)\text{P}\left( \dfrac{\text{I}}{\text{A}} \right)}{\text{P}\left( \text{I} \right)}$ thus answer will be incorrect and students will get very confused where to put which value. Also, remember to multiply the number of processes repeated, if forget to put in the calculation part then the whole answer can be changed. So, do not make this type of mistake.