Question

A bag contains four pair of balls of four distinct colours. If four balls are picked at random (without replacement), then what is the probability that there is at least one pair among them have the same colour?
A. \[\dfrac{1}{{7!}}\]
B. \[\dfrac{8}{{35}}\]
C. \[\dfrac{{19}}{{35}}\]
D. \[\dfrac{{27}}{{35}}\]

Answer 1

Hint: In the given question, a bag has four pairs of balls of four distinct colors. We have to find the probability that there is at least one pair among them has the same color. The total number of balls present in the bag is 8 and we have to choose four balls so that none of the pairs has the same color. We will apply the probability formula to compute the required probability.
Formula Used:
Probability Formula:
The probability of an event \[E\] is: \[P\left( E \right) = \dfrac{{The\, number\, of\, favourable \,outcomes}}{{Total\, number\, of \,outcomes}}\]
\[P\left( E \right) + P\left( {E'} \right) = 1\]
The formula of combination: \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\]
Complete step by step solution:
It is given that a bag contains four pairs of balls of four distinct colors.
From the given data,
The total number of balls in the bag is 4 pairs.
\[ \Rightarrow \] Total number of balls in the bag is 8 balls.
We have to choose 4 balls randomly.
The total outcomes, i.e., total number of ways to select the four balls at random is:
 \[{}^8{C_4} = \dfrac{{8!}}{{4!\left( {8 - 4} \right)!}}\]
\[ \Rightarrow \]\[{}^8{C_4} = \dfrac{{8!}}{{4! \cdot 4!}}\]
\[ \Rightarrow \]\[{}^8{C_4} = \dfrac{{8 \times 7 \times 6 \times 5}}{{1 \times 2 \times 3 \times 4}}\]
\[ \Rightarrow \]\[{}^8{C_4} = 70\]

In the bag, each pair of balls has the same color.
The number of ways to select 4 balls, in which none of the balls have the same colour is:
\[{}^2{C_1} \cdot {}^2{C_1} \cdot {}^2{C_1} \cdot {}^2{C_1} = \dfrac{{2!}}{{1!\left( {2 - 1} \right)!}} \cdot \dfrac{{2!}}{{1!\left( {2 - 1} \right)!}} \cdot \dfrac{{2!}}{{1!\left( {2 - 1} \right)!}} \cdot \dfrac{{2!}}{{1!\left( {2 - 1} \right)!}}\]
\[ \Rightarrow \]\[{}^2{C_1} \cdot {}^2{C_1} \cdot {}^2{C_1} \cdot {}^2{C_1} = 2 \cdot 2 \cdot 2 \cdot 2\]
\[ \Rightarrow \]\[{}^2{C_1} \cdot {}^2{C_1} \cdot {}^2{C_1} \cdot {}^2{C_1} = 16\]

Apply the probability formula to calculate the probability of selecting balls, which does not have the same colour.
\[P\left( E \right) = \dfrac{{16}}{{70}}\]
\[ \Rightarrow \]\[P\left( E \right) = \dfrac{8}{{35}}\]

Therefore, the required probability is:
required probability \[ = 1 - P\left( E \right)\]
\[ \Rightarrow \] required probability \[ = 1 - \dfrac{8}{{35}}\]
\[ \Rightarrow \] required probability \[ = \dfrac{{35 - 8}}{{35}}\]
\[ \Rightarrow \] required probability \[ = \dfrac{{27}}{{35}}\]

Hence the correct option is D.
Note: Probability means how likely something is to happen. The probability of an event lies from 0 to 1.
The sum of the probability of an event and the probability of its complement is always 1.