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) $\dfrac{21}{46}$
b) $\dfrac{25}{117}$
c) $\dfrac{1}{50}$
d) $\dfrac{3}{25}$
Answer
673.5k+ views
Hint: We will first find the number of ways of selecting 3 out of 25, then we will find the number of ways of selecting 1 girl out of 10 girl and 2 boys out of 15, then we will find probability required as $\dfrac{\text{Number of ways of selecting 2 boys and 1 girl}}{\text{Total Sample}}$, also we will use the formula ${}^{n}{{c}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.
Complete step-by-step answer:
It is given in the question that there are 15 boys and 10 girls, three students are selected at random, then we have to find the probability that 1 girl and 2 boys are selected at random.
Let us assume a sample space (S), where the number of ways of selecting 3 students out of 25 is $n\left( s \right)={}^{25}{{C}_{3}}$. Using the formula ${}^{n}{{c}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$, we get
${}^{25}{{C}_{3}}=\dfrac{25!}{3!\left( 25-3 \right)!}=\dfrac{25\times 24\times 23\times 22!}{3!\times 22!}$, solving further, we get
\[{}^{25}{{C}_{3}}=\dfrac{25\times 24\times 23}{3\times 2\times 1}=\dfrac{25\times 24\times 23}{6}=25\times 4\times 23=2300\].
Now, the number of ways of selecting 1 girl out of 10 girls will be $={}^{10}{{C}_{1}}=10$ ( using the formula ${}^{n}{{c}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$).
Similarly, the number of ways of selecting 2 boys out of 15 will be $={}^{15}{{C}_{2}}=\dfrac{15!}{2!\left( 15-2 \right)!}=\dfrac{15\times 14}{2}=105$. Thus, the required probability that 1 girl and 2 boys are selected at random is given by $P=\dfrac{\left( \text{Number of ways of selecting 1 girl} \right)\times \left( \text{Number of ways of selecting 2 boys} \right)}{\text{Number of ways of selecting 3 students out of 25}}$, on putting the values, we get
$P=\dfrac{10\times 105}{2300}=\dfrac{1050}{2300}=\dfrac{105}{230}=\dfrac{21}{46}$.
Therefore, option a) is the correct answer.
Note: Students may find a number of ways of selecting boys among all 25 students and similarly, a number of ways of selecting girls among all 25 students which is not correct at all. We have to find the number of ways of selecting boys among 15 and the number of ways of selecting girls among 10 and not among 25.
Complete step-by-step answer:
It is given in the question that there are 15 boys and 10 girls, three students are selected at random, then we have to find the probability that 1 girl and 2 boys are selected at random.
Let us assume a sample space (S), where the number of ways of selecting 3 students out of 25 is $n\left( s \right)={}^{25}{{C}_{3}}$. Using the formula ${}^{n}{{c}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$, we get
${}^{25}{{C}_{3}}=\dfrac{25!}{3!\left( 25-3 \right)!}=\dfrac{25\times 24\times 23\times 22!}{3!\times 22!}$, solving further, we get
\[{}^{25}{{C}_{3}}=\dfrac{25\times 24\times 23}{3\times 2\times 1}=\dfrac{25\times 24\times 23}{6}=25\times 4\times 23=2300\].
Now, the number of ways of selecting 1 girl out of 10 girls will be $={}^{10}{{C}_{1}}=10$ ( using the formula ${}^{n}{{c}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$).
Similarly, the number of ways of selecting 2 boys out of 15 will be $={}^{15}{{C}_{2}}=\dfrac{15!}{2!\left( 15-2 \right)!}=\dfrac{15\times 14}{2}=105$. Thus, the required probability that 1 girl and 2 boys are selected at random is given by $P=\dfrac{\left( \text{Number of ways of selecting 1 girl} \right)\times \left( \text{Number of ways of selecting 2 boys} \right)}{\text{Number of ways of selecting 3 students out of 25}}$, on putting the values, we get
$P=\dfrac{10\times 105}{2300}=\dfrac{1050}{2300}=\dfrac{105}{230}=\dfrac{21}{46}$.
Therefore, option a) is the correct answer.
Note: Students may find a number of ways of selecting boys among all 25 students and similarly, a number of ways of selecting girls among all 25 students which is not correct at all. We have to find the number of ways of selecting boys among 15 and the number of ways of selecting girls among 10 and not among 25.
Recently Updated Pages
Which of the following graphs shows the variation of class 12 physics CBSE

Draw a labelled diagram of the human male reproductive class 12 biology CBSE

Describe the experiment to compare the emf of two cells class 12 physics CBSE

What is standard hydrogen electrode

What is conventional current and electric current class 12 physics CBSE

2Bromopentane is treated with an alcoholic KOH solution class 12 chemistry CBSE

Trending doubts
Draw a labelled sketch of the human eye class 12 physics CBSE

Which are the Top 10 Largest Countries of the World?

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Draw ray diagrams each showing i myopic eye and ii class 12 physics CBSE

Which is the correct genotypic ratio of mendel dihybrid class 12 biology CBSE

What is the Full Form of PVC, PET, HDPE, LDPE, PP and PS ?

