Question

From 4 red balls, 2 white balls and 4 black balls, 4 balls are selected. The probability of getting 2 red balls is
A) $\dfrac{7}{{21}}$
B) $\dfrac{8}{{21}}$
C) $\dfrac{9}{{21}}$
D) $\dfrac{{10}}{{21}}$

Answer 1

We will find the total outcomes and the favourable outcomes using the formula for combination. Then we will find the probability by using the formula for probability. We will cancel out the common factors (if there are any) and finally choose the correct option.

Formulas used: We will use the following formulas:

1. The number of ways of selecting $r$ objects from a set of $n$ objects is ${}^n{C_r}$
2. The formula for ${}^n{C_r}$ is ${}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$
3. The formula for probability (P) is given by $P =$ number of favourable outcomes $\div$ total number of outcomes.

Complete step by step solution:
We know that there are 4 red balls, 2 white balls and 4 black balls. We will find the total number of balls:
$4 + 2 + 4 = 10$
There are 10 balls in total. We will find the total number of ways in which 4 balls can be selected from 10 balls. We will substitute 4 for $r$ and 10 for $n$ in ${}^n{C_r}$:
${}^{10}{C_4} = \dfrac{{10!}}{{\left( {10 - 4} \right)!4!}}$
Computing the factorial, we get
$\Rightarrow {}^{10}{C_4} = \dfrac{{10 \times 9 \times 8 \times 7 \times 6!}}{{6!4!}}$
Simplifying the expression, we get
$\begin{array}{l} \Rightarrow {}^{10}{C_4} = \dfrac{{10 \times 9 \times 8 \times 7}}{{24}}\\\ \Rightarrow {}^{10}{C_4} = 210\end{array}$
There are 210 ways in which 4 balls can be selected from 10 balls. So, the total number of outcomes is 210.
We want to select exactly 2 red balls. We will substitute 2 for $r$ and 4 for $n$ in ${}^n{C_r}$to find the number of ways in which 2 red balls can be selected out of 4 red balls.
Therefore,
${}^4{C_2} = \dfrac{{4!}}{{\left( {4 - 2} \right)!2!}}$
Computing the factorial, we get
$\begin{array}{l} \Rightarrow {}^4{C_2} = \dfrac{{24}}{{2 \times 2}}\\\ \Rightarrow {}^4{C_2} = 6\end{array}$
There are 6 ways in which exactly 2 red balls can be selected from 4 red balls.
We have to select 4 balls out of which 2 should be red balls. The rest 2 can be white or black balls. We know that there are 2 white balls and 4 black balls. We will find the number of ways of selecting 2 balls from 6 balls.
Substituting 2 for and 6 for $n$ in ${}^n{C_r}$, we get
${}^6{C_2} = \dfrac{{6!}}{{\left( {6 - 2} \right)!2!}}$
Computing the factorial, we get
$\begin{array}{l} \Rightarrow {}^6{C_2} = \dfrac{{720}}{{24 \times 2}}\\\ \Rightarrow {}^6{C_2} = 15\end{array}$
The total number of ways of selecting 4 balls such that exactly 2 red balls are selected is:
$6 \times 15 = 90$
So, the number of favourable outcomes is 90.
We will find the probability by substituting 90 in the numerator and 210 in the denominator in the formula for probability.
$P = \dfrac{{90}}{{210}}$
Breaking the terms, we get
$\begin{array}{l} \Rightarrow P = \dfrac{{9 \times 10}}{{21 \times 10}}\\\ \Rightarrow P = \dfrac{9}{{21}}\end{array}$

$\therefore$ Option C is the correct option.

Note:
We must know that the formula for combination (${}^n{C_r}$) is used in cases where the order of selection does not matter whereas in cases where the order of selection also matters, we should use the formula for permutations. The formula for permutations is given by ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$ where $n$ is the total number of objects and $r$ is the number of objects to be selected.