Question

In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected is:
(a) $\dfrac{25}{46}$
(b) $\dfrac{15}{46}$
(c) $\dfrac{21}{46}$
(d) $\dfrac{22}{46}$

Answer 1

Hint: To solve this question, we will first find out the number of ways in which students can be selected. Let this value be y. Now, we will find the number of ways in which we can select 1 girl and 2 boys. Let this value be x. Now, we will find the probability with the help of the formula given below:
$\text{Probability}=\dfrac{\text{Favorable Outcomes}}{\text{Total Number of Outcomes}}$

Complete step-by-step answer:
In this question, we have to determine the probability that the selected students are 2 boys and 1 girl. For this, we will first find out the total number of ways in which we can select the students. Now, we are given that there are 15 boys and 10 girls in the class. Thus, the total students in that class are 15 + 10 = 25. It is also given that we have to select any three students. Thus, the number of ways in which we can select any three students from 25 students $={{\text{ }}^{25}}{{C}_{3}}.$
Now, we will find the number of ways in which we can select 2 boys and 1 girl. The number of ways in which we can select 2 boys from 15 boys $={{\text{ }}^{15}}{{C}_{2}}.$ The number of ways in which we can select 1 girl from 10 girls $={{\text{ }}^{10}}{{C}_{1}}.$ Hence, the total ways of doing this $={{\text{ }}^{15}}{{C}_{2}}\times {{\text{ }}^{10}}{{C}_{1}}.$
Now, we will find the probability. Probability of any event is given by the formula:
$\text{Probability}=\dfrac{\text{Favorable Outcomes}}{\text{Total Number of Outcomes}}$
In our class, the total number of outcomes $={{\text{ }}^{25}}{{C}_{3}}$ and favorable outcomes $={{\text{ }}^{15}}{{C}_{2}}\times {{\text{ }}^{10}}{{C}_{1}}$
Thus,
$\text{Probability }=\dfrac{^{15}{{C}_{2}}\times {{\text{ }}^{10}}{{C}_{1}}}{^{25}{{C}_{3}}}$
Here, we are going to use the formula for $^{n}{{C}_{r}}$ which is shown:
$^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}=\dfrac{n\left( n-1 \right)\left( n-2 \right)....3.2.1}{\left[ 1.2.3.....\left( n-r \right) \right]\left[ 1.2.3.....r \right]}$
$^{n}{{C}_{r}}=\dfrac{n\left( n-1 \right)\left( n-2 \right)....\left( n-r+1 \right)}{1.2.3......r}$
$^{n}{{C}_{r}}=\dfrac{n\left( n-1 \right)\left( n-2 \right)....\left( n-r+1 \right)}{r!}$
Thus, we will get,
$\text{Probability }=\dfrac{\dfrac{15\times 14}{2}\times \dfrac{10}{1}}{\dfrac{25\times 24\times 23}{6}}=\dfrac{21}{46}$
Hence, the option (c) is the right answer.

Note: We can also find favorable outcomes in the given question by first finding the total outcomes and then subtracting the non-favorable outcomes from it. Thus the probability is given by:
$\text{Probability }=\dfrac{^{25}{{C}_{3}}-\left( ^{15}{{C}_{1}}\times {{\text{ }}^{10}}{{C}_{2}}{{+}^{15}}{{C}_{3}}\times {{\text{ }}^{10}}{{C}_{0}}{{+}^{15}}{{C}_{0}}\times {{\text{ }}^{10}}{{C}_{3}} \right)}{^{25}{{C}_{3}}}$
$\text{Probability }=1-\dfrac{\left( ^{15}{{C}_{1}}\times {{\text{ }}^{10}}{{C}_{2}} \right)+\left( ^{15}{{C}_{3}}\times {{\text{ }}^{10}}{{C}_{0}} \right)+\left( ^{15}{{C}_{0}}\times {{\text{ }}^{10}}{{C}_{3}} \right)}{^{25}{{C}_{3}}}$
$\text{Probability }=1-\dfrac{\left( 15\times 45 \right)+\left( 455\times 1 \right)+\left( 120\times 1 \right)}{2300}$
$\text{Probability }=1-\dfrac{675+455+120}{2300}$
$\text{Probability }=1-\dfrac{1250}{2300}$
$\text{Probability }=1-\dfrac{25}{46}$
$\text{Probability }=\dfrac{21}{46}$