Question

Let ${u_1}$ and ${u_2}$ be two urns such that ${u_1}$ contains 3 white, 2 red balls and ${u_2}$ contains only 1 white ball. A fair coin is tossed. If the head appears, then 1 ball is drawn at random from urn ${u_1}$ and put into ${u_2}$. However, if the tail appears, then 2 balls are drawn at random from ${u_1}$ and put into ${u_2}$. Now, 1 ball is drawn at random from ${u_2}$. Then, probability of the drawn ball from ${u_2}$ being white is
A) $\dfrac{{13}}{{30}}$
B) $\dfrac{{23}}{{30}}$
C) $\dfrac{{19}}{{30}}$
D) $\dfrac{{11}}{{30}}$

Answer 1

Hint : In this question, we need to determine the probability of the drawn ball from ${u_2}$ being white. For this we will use the basic definition of the probability along with the combinations of the similar objects.

Complete step-by-step answer :
Urn ${u_1}$ contains 3 white and 2 red balls, so the total number of balls in urn ${u_1}$ is 5.
Urn ${u_2}$ contains 1 white ball
Now it is said that a coin is tossed and if Head appear in the coin then 1 ball is drawn at random from urn ${u_1}$ and put into ${u_2}$,
So the probability of drawing the ball from urn ${u_1}$ if the ball drawn at random is white
$\Rightarrow {P_1} =\left( {\dfrac{1}{2}}\right)\left( {\dfrac{3}{5}}\right)\left( {\dfrac{2}{2}}\right) =\dfrac{3}{{10}}$
Now again if the ball drawn at random from urn ${u_1}$ is red ball, so the probability becomes
$\Rightarrow {P_2} =\left( {\dfrac{1}{2}}\right)\left( {\dfrac{2}{5}}\right)\left( {\dfrac{1}{2}}\right) =\dfrac{1}{{10}}$
Now it is said that when the coin was tossed and if the coin would have showed with tail,
So the probability of drawing the ball from urn ${u_1}$ if the ball drawn at random is 2 white,
$\Rightarrow {P_3} =\left( {\dfrac{1}{2}}\right)\left( {\dfrac{{{}^3{C_2}}}{{{}^5{C_2}}}}\right)\left( {\dfrac{3}{3}}\right) =\left( {\dfrac{1}{2}}\right)\left( {\dfrac{3}{{\dfrac{{5\times 4}}{2}}}}\right)\left( {\dfrac{3}{3}}\right) =\dfrac{3}{{20}}$
Now again if the ball drawn at random from urn ${u_1}$ is 1 red ball and 1 white, so the probability becomes
$\Rightarrow {P_4} =\left( {\dfrac{1}{2}}\right)\left( {\dfrac{{{}^3{C_1} \times {}^2{C_1}}}{{{}^5{C_2}}}}\right)\left( {\dfrac{2}{3}}\right) =\left( {\dfrac{1}{2}}\right)\left( {\dfrac{{3 \times 2}}{{\dfrac{{5\times 4}}{2}}}}\right)\left( {\dfrac{2}{3}}\right) =\dfrac{1}{5}$
Now if the ball drawn at random from urn ${u_1}$ is 2 red ball, so the probability
$\Rightarrow {P_5} =\left( {\dfrac{1}{2}}\right)\left( {\dfrac{{{}^2{C_2}}}{{{}^5{C_2}}}}\right)\left( {\dfrac{1}{3}}\right) =\left( {\dfrac{1}{2}}\right)\left( {\dfrac{1}{{\dfrac{{5\times 4}}{2}}}}\right)\left( {\dfrac{2}{3}}\right) =\dfrac{1}{{60}}$
So the total probability of the drawing ball from ${u_2}$

$$\Rightarrow P = {P_1} + {P_2} + {P_3} + {P_4} + {P_5}\\\ =\dfrac{3}{{10}} +\dfrac{1}{{10}} +\dfrac{3}{{20}} +\dfrac{1}{5} +\dfrac{1}{{60}}\\\ =\dfrac{{45}}{{60}}\\\ =\dfrac{{23}}{{60}}\\\$$

Hence the probability of the drawn ball from ${u_2}$ being white is$=\dfrac{{23}}{{60}}$
So, the correct answer is “Option B”.

Note :
${}^n{C_r}$ is the mathematical representation of the combination which is a method of selection of some items or all of the items from a set without considering the sequence of selection whereas in the case of permutation which is the method of arrangements of items of a set the sequence is considered represented as ${}^n{P_r}$ .
${}^n{C_r} =\dfrac{{n!}}{{\left( {n - r}\right)!r!}}$
${}^n{P_r} =\dfrac{{n!}}{{\left( {n - r}\right)!}}$