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 \[{}^n{P_r}\] .
\[{}^n{C_r} =\dfrac{{n!}}{{\left( {n - r}\right)!r!}}\]
\[{}^n{P_r} =\dfrac{{n!}}{{\left( {n - r}\right)!}}\]