Question

In a group of students, there are 3 boys and 3 girls. Four students are to be selected randomly from the group. Find the probability that either 3 boys and 1 girl or 3 girls and 1 boy are selected.

Answer 1

Hint: we use the concepts of probability to solve the given question. We start to solve the given question by calculating all the favorable outcomes for the event. Then, we divide all the favorable outcomes by total outcomes to get the desired result.

Complete step by step solution:
We are asked to calculate the probability that either 3 boys and 1 girl or 3 girls and 1 boy are selected from a group of 3 boys and 3 girls. We will be solving the given question using the concept of probability.
From the above, we can see that there can be two possible cases in this situation, the first case involves the selection of 3 boys and 1 girl.
The number of ways 3 boys can be selected out of a group of 3 boys and a girl out of a group of 3 girls is found out using combinations.
Combinations usually are the number of ways the objects can be selected without replacement. It is usually given by $^{n}{{C}_{r}}$
Here,
n = no of things
r = no of things taken at the time.
The combination of n things taken r at a time or $^{n}{{C}_{r}}$ is given by the formula
 ${{\Rightarrow }^{n}}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$
The number of ways 3 boys can be selected out of a group of 3 boys is given by $^{3}{{C}_{3}}$
The number of ways a girl can be selected out of a group of 3 girls is given by $^{3}{{C}_{1}}$
The number of ways 3 boys can be selected out of a group of 3 boys and a girl can be selected out of a group of 3 girls is given as follows,
${{\Rightarrow }^{3}}{{C}_{3}}{{\times }^{3}}{{C}_{1}}$
$\Rightarrow \dfrac{3!}{3!\left( 3-3 \right)!}\times \dfrac{3!}{1!\left( 3-1 \right)!}$
Simplifying the above expression, we get,
$\Rightarrow 3$
The second case involves selection of 1 boy and 3 girls.
The number of ways a boy can be selected out of a group of 3 boys is given by $^{3}{{C}_{1}}$
The number of ways 3 girls can be selected out of a group of 3 girls is given by $^{3}{{C}_{3}}$
The number of ways a boy can be selected out of a group of 3 boys and 3 girls can be selected out of a group of 3 girls is given as follows,
${{\Rightarrow }^{3}}{{C}_{1}}{{\times }^{3}}{{C}_{3}}$
$\Rightarrow \dfrac{3!}{1!\left( 3-1 \right)!}\times \dfrac{3!}{3!\left( 3-3 \right)!}$
Simplifying the above expression, we get,
$\Rightarrow 3$
The favorable outcomes from the given question is the selection of either 3 boys and 1 girl or 3 girls and 1 boy from the group.
$\Rightarrow \text{favorable outcomes = 3+3}$
$\therefore \text{Favorable outcomes = 6}$
The total number of possible outcomes from the given question is the selection of 4 students from a group of 6 students(3 boys and 3 girls).
The number of ways 4 students can be selected out of a group of 6 students is given by $^{6}{{C}_{4}}$
$\Rightarrow \text{Total outcomes =}{{\text{ }}^{6}}{{C}_{4}}$
$\Rightarrow \text{Total outcomes = }\left( \dfrac{6!}{4!\left( 6-4 \right)!} \right)$
$\Rightarrow \text{Total outcomes = }\left( \dfrac{6\times 5\times 4\times 3\times 2\times 1}{4\times 3\times 2\times 1\times 2} \right)$
$\Rightarrow \text{Total outcomes = }\left( \dfrac{6\times 5}{2} \right)$
$\Rightarrow \text{Total outcomes = }\dfrac{30}{2}$
$\therefore \text{Total outcomes = 15}$
We know that,
$\Rightarrow \text{probability = }\dfrac{\text{number of favorable outcomes}}{\text{Total outcomes}}$
Substituting the values, we get,
$\Rightarrow probability=\dfrac{6}{15}$
Simplifying the above equation, we get,
$\therefore \text{probability = }\dfrac{2}{5}$

The required probability is $\dfrac{2}{5}$

Note: We need to know the basic concepts and definitions of combinations to solve this problem. Combinations usually are the number of ways the objects can be selected without replacement. The factorial of a number n should only go down to 1 and not zero. Common mistakes like double counting, confusion of values should be avoided to get precise results.