Question

A box ‘A’ contains 2 white, 3 red and 2 black balls. Another box ‘B’ contains 4 white, 2 red and 3 black balls. If two balls are drawn at random, without replacement, from a randomly selected box and one ball turns out to be white while the other ball turns out to be red, then the probability that both balls are drawn from box ‘B’ is
A. \[\dfrac{7}{{16}}\]
B. \[\dfrac{9}{{32}}\]
C. \[\dfrac{7}{8}\]
D. \[\dfrac{9}{{16}}\]

Answer 1

Hint: Here we will proceed by selecting the two balls from both the boxes separately and then consider the event that the balls are drawn from box ‘B’. Further find the number of favourable and total outcomes of the considered the event to get the required solution.

Complete step-by-step answer:
Given that in box ‘A’ there are 2 white, 3 red and 2 black balls. So, there are a total of 7 balls in box ‘A’.
Also given that in box ‘B’ there are 4 white, 2 red and 3 black balls. So, there are a total of 9 balls in box ‘B’.
Given that two balls are drawn at randomly without replacement from a randomly selected box. So, there are 2 ways of selecting a box randomly i.e., box ‘A’ and box ‘B’.
Then we have to draw two balls from the selected box in which the first ball turns out to be a white ball and then second ball turns out to be a red ball.
Now, consider the that two balls are drawn from box ‘A’.
The number of ways selecting a white ball first and a red ball second from 7 balls \[ = \dfrac{{{}^2{C_1} \times {}^3{C_1}}}{{{}^7{C_2}}}\]
Then, consider the that two balls are drawn from box ‘B’.
The number of ways selecting a white ball first and a red ball second from 9 balls \[ = \dfrac{{{}^4{C_1} \times {}^2{C_1}}}{{{}^9{C_2}}}\]
We know that the probability of an event \[E\] is given by \[P\left( E \right) = \dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes}}}}\]
Let \[E\] be the event that getting a white ball first and a ed ball from the box ‘B’ without replacement.
So, the number of favourable outcomes \[ = \dfrac{{{}^4{C_1} \times {}^2{C_1}}}{{{}^9{C_2}}}\]
And the total number of outcomes \[ = \dfrac{{{}^4{C_1} \times {}^2{C_1}}}{{{}^9{C_2}}} + \dfrac{{{}^2{C_1} \times {}^3{C_1}}}{{{}^7{C_2}}}\]
Hence, the probability of event \[E\] is given by
\[
   \Rightarrow P\left( E \right) = \dfrac{{\dfrac{{{}^4{C_1} \times {}^2{C_1}}}{{{}^9{C_2}}}}}{{\dfrac{{{}^4{C_1} \times {}^2{C_1}}}{{{}^9{C_2}}} + \dfrac{{{}^2{C_1} \times {}^3{C_1}}}{{{}^7{C_2}}}}} \\
   \Rightarrow P\left( E \right) = \dfrac{{\dfrac{{4 \times 2}}{{36}}}}{{\dfrac{{4 \times 2}}{{36}} + \dfrac{{2 \times 3}}{{21}}}} \\
\]
Cancelling, the common terms we have
\[
   \Rightarrow P\left( E \right) = \dfrac{{\dfrac{2}{9}}}{{\dfrac{{7 \times 2 + 9 \times 2}}{{9 \times 7}}}} \\
   \Rightarrow P\left( E \right) = \dfrac{{\dfrac{2}{9}}}{{\dfrac{{14 + 18}}{{9 \times 7}}}} \\
   \Rightarrow P\left( E \right) = \dfrac{{\dfrac{2}{9}}}{{\dfrac{{32}}{{9 \times 7}}}} \\
\]
Simplifying further, we get
\[
   \Rightarrow P\left( E \right) = \dfrac{{2 \times 7}}{{32}} \\
  \therefore P\left( E \right) = \dfrac{7}{{16}} \\
\]
Thus, the correct option is A. \[\dfrac{7}{{16}}\]

So, the correct answer is “Option A”.

Note: The probability of an event is always lying between 0 and 1 i.e., \[0 \leqslant P\left( E \right) \leqslant 1\]. The probability of an event \[E\] is given by \[P\left( E \right) = \dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes}}}}\]. Here we used the formula \[{}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}\] to simplify the equations obtained.