Question

In a class there are $10$ boys and $8$ girls. When $3$ students are selected at random, the probability that $2$ girls and $1$ boy are selected is
A. $\dfrac{{35}}{{102}}$
B. $\dfrac{{15}}{{102}}$
C. $\dfrac{{55}}{{102}}$
D. $\dfrac{{25}}{{102}}$

Answer 1

This problem can be solved easily by using the probability formula.
The probability of individual disjoint events can be found by using the combination formula.
Then by the AND rule, we will find the probability of those two events together.
Formula: If $n$is the total number of outcome and if $r$ is the favorable outcome then the probability of getting $r$ in $n$ $=$ $n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$
Probability of an event = Number of favorable outcomes$/$ total number of outcomes

Complete step by step answer:
It is given that there are $10$ boys and $8$ girls in the class. We aim to find the probability of selecting $2$ girls and $1$boys in a random selection of $3$ students.
We can see that the two disjoint events are getting a girl and getting a boy in selecting three students. So first we have to find the probability of these two disjoint events.
Let us first find the probability of the event of selecting two girls.
We know that there are eight girls in total. Therefore, the probability of selecting two girls from a total number of girls will be $^8{C_2}$
Let us find the value of this combination using the formula $n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$.
$^8{C_2} = \dfrac{{8!}}{{2!(8 - 2)!}} = \dfrac{{8!}}{{2!(6!)}}$
Let’s simplify this.
$= \dfrac{{8 \times 7 \times 6!}}{{2! \times 6!}}$
On further simplification we get
$^8{C_2} = \dfrac{{8 \times 7}}{2}$
Let’s keep this value as it is.
Now let’s find the probability next event. That is getting one boy at a random selection.
We know that there are $10$ boys in total. Therefore, the probability of selecting one boy from the total number of boys will be $^{10}{C_1}$.
Let us find the value of this combination using the formula $n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$.
$^{10}{C_1} = \dfrac{{10!}}{{1!(10 - 1)!}}$
Let us simplify this.
$^{10}{C_1} = \dfrac{{10 \times 9!}}{{1 \times 9!}}$
On further simplification we get
$^{10}{C_1} = 10$
Thus, the probability of getting one boy on selection is $10$.
Now let us find the probability of selecting three students from a whole class.
We know that the total number of students in a class is $10$boys + $$$$8girls $= 18$students.
Therefore, selecting three students at random can be found by the combination $^{18}{C_3}$.
Let us simplify this to get the value.
$^{18}{C_3} = \dfrac{{18!}}{{3!(18 - 3)!}} = \dfrac{{18 \times 17 \times 16 \times 15!}}{{3 \times 2 \times 15!}}$
On simplifying this we get
$^{18}{C_3} = \dfrac{{18 \times 17 \times 16}}{{3 \times 2}}$
Let’s keep this value as it is.
Now we have to find the probability of getting two girls and one boy at a random selection.
Using the formula, the Probability of an event$=$ Number of favorable outcomes$/$ total number of outcomes.
Then, the probability of getting two girls and one boy $=$ the product of the probability of getting two girls and the probability of getting one boy divided by the probability of selecting three students at a random selection.
That is the probability of getting two girls and one boy $=$ $\dfrac{{^8{C_2}{ \times ^{10}}{C_1}}}{{^{18}{C_3}}}$
We already know the values of $^8{C_2}{,^{10}}{C_1}{\& ^{18}}{C_3}$ let’s substitute them in the above formula.
The probability of getting two girls and one boy $=$ $\dfrac{{\left( {\dfrac{{8 \times 7}}{2}} \right) \times 10}}{{\dfrac{{18 \times 17 \times 16}}{{3 \times 2}}}}$
On simplifying this we get
$= \dfrac{{8 \times 7 \times 5 \times 3 \times 2}}{{18 \times 17 \times 16}}$
On further simplification we get
$= \dfrac{{35}}{{102}}$
Thus, the probability of getting two girls and one boy at a random selection is $\dfrac{{35}}{{102}}$.
So, the correct answer is “Option A”.

Note: We cannot use conditional probability formulas here since the given two events are disjoint events.
So, we need to use a combination formula to find the probability of two given disjoint events.
The combination gives the possible outcomes of the specified events.