Question

A bag contains 6 white and 4 black balls. 2 balls are drawn at random. Find the probability that they are of the same colour.

Answer 1

Hint: In this question, we are given 10 balls of 2 different colours and we have been asked to find the probability of drawing the balls of the same colour. It is also given that only 2 balls are drawn at once. First, find the number of ways of drawing 2 balls out of 10. Then, find the number of ways of drawing 2 same coloured balls – either 2 white or 2 black. Then put them in the formula of probability and simplify to get your required answer.

Formula used:
1) Probability = $\dfrac{{{\text{Number of favourable items}}}}{{{\text{Number of total items}}}}$
2) $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$

Complete step-by-step solution:
We are given 6 white and 4 black balls and we have been asked to find the probability of drawing 2 balls of the same colour.
At first, we will find the number of ways of drawing 2 balls out of 10.
2 balls can be drawn out of 10 balls in the following number of ways –
${ \Rightarrow ^{10}}{C_2} = \dfrac{{10!}}{{2! \times \left( {10 - 2} \right)!}}$ …. (using $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$)
${ \Rightarrow ^{10}}{C_2} = \dfrac{{10 \times 9 \times 8!}}{{2! \times 8!}}$
${ \Rightarrow ^{10}}{C_2} = 45$
Therefore, total outcomes are 45.
Now, we will find a number of favourable items.
Favourable items = n(2 white balls appear) + n(2 black balls appear)
Favourable items = $^6{C_2}{ + ^4}{C_2}$
On simplifying we will get,
Favourable items = $\dfrac{{6!}}{{2! \times 4!}} + \dfrac{{4!}}{{2! \times 2!}}$
Favourable items = $\dfrac{{6 \times 5 \times 4!}}{{2! \times 4!}} + \dfrac{{4 \times 3 \times 2!}}{{2! \times 2!}}$
Favourable items = $15 + 6 = 21$
Now, we will put both of them in the formula,
$ \Rightarrow $ Probability = $\dfrac{{{\text{Number of favourable items}}}}{{{\text{Number of total items}}}}$
$ \Rightarrow $ Probability = $\dfrac{{21}}{{45}}$$ = \dfrac{7}{{15}}$

Hence, the probability that the balls are of the same colour is $\dfrac{7}{{15}}$.

Note: Here, we have to calculate the probability of drawing 2 balls either out of $6$ white or 4 black balls. We used the word ‘or’ in this statement. In such cases, we add the two cases.
In certain cases, we will have to find the probability of drawing 2 balls out of $6$ white and 4 black balls. We used the word ‘and’ in this statement. In such cases, we multiply the two cases.