Question

A bag contains $3$ red, $4$ white and $5$ black balls. Three balls are drawn at random. The probability of being their different colours is
A. $\dfrac{3}{{11}}$
B. $\dfrac{2}{{11}}$
C. $\dfrac{8}{{11}}$
D. None of these

Answer 1

We can see that in this question we have to find the probability of selecting the balls from different bags. So this applies to the combination. So we will first find the total outcomes and the favourable outcomes through the combination formula and then we apply the probability formula. We know the formula of probability i.e. $\dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$.

Complete step by step answer:
We know the formula for the number of ways for selecting $r$ things from $n$ group of people. The formula of combination is given as;
$^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ .
Here we have total number of balls;
$3 + 4 + 5 = 12$
And we have to select three balls from a total of $12$ balls. So we have :
$n = 12,r = 3$
We can write this expression as
$^{12}{C_3}$
Now by putting the value in the formula we can write:
$\dfrac{{n!}}{{r!\left( {n - r} \right)!}} = \dfrac{{12!}}{{3!(12 - 3)!}}$
We can simplify the factorial values and it gives us
$\dfrac{{12 \times 11 \times 10 \times 9!}}{{3 \times 2 \times 9!}}$

Now it gives us the value:
$4 \times 11 \times 5 = 220$
Now we will calculate all the favourable cases of this combination:
Since we have to find the probability of all three balls of different colours, we will calculate the combination of one ball from each.
First we have a total number of red balls $3$.
And we have to select one red from a total $3$ balls . Here we have: $n = 3,r = 1$.
Now by putting the value in the formula we can write:
$\dfrac{{n!}}{{r!\left( {n - r} \right)!}} = \dfrac{{3!}}{{1!(3 - 1)!}}$
On simplifying it gives us :
$\dfrac{{3 \times 2!}}{{2!}} = 3$
Second we have a total number of white balls $4$.

And we have to select one white ball from a total $4$ balls . So we have the values: $n = 4,r = 1$.
Now by putting the value in the formula we can write:
$\dfrac{{n!}}{{r!\left( {n - r} \right)!}} = \dfrac{{4!}}{{1!(4 - 1)!}}$
On simplifying it gives us the values:
$\dfrac{{4 \times 3!}}{{3!}} = 4$
Now we have a total number of black balls $5$. And we have to select one black from a total $5$ black balls . So we have :
$n = 5,r = 1$
Similarly as above by putting the value in the formula we can write:
$\dfrac{{n!}}{{r!\left( {n - r} \right)!}} = \dfrac{{5!}}{{1!(5 - 1)!}}$
On further simplifying it gives us the value :
$\dfrac{{5 \times 4!}}{{4!}} = 5$

We know the formula of probability i.e. $\dfrac{\text{No. of favourable outcomes}}{\text{Total number of outcomes}}$.
We will apply this now to solve further problems.
Now we have the total number of favourable outcomes i.e.
$3 \times 4 \times 5$
And we have total number of outcomes
$220$
Now we can put the values in the formula and it gives:
$\dfrac{{3 \times 4 \times 5}}{{220}} = \dfrac{{60}}{{220}}$
We can simplify the value, and it gives: $\dfrac{3}{{11}}$.

Hence the correct option is A.

Note: We should know that probability is the prediction of a particular outcome of a random event. It is a set of all the possible outcomes for a random experiment. We should note that if in the question it is given that we have to arrange the balls in a specific order, then we will use the formula of Permutation.