Subject: General Aptitude · Chapter: Probability · Exam: · Marks: · Difficulty:
In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that the selected students are 2 boys and 1 girl, is:
(a)$\frac{21}{46}$
(b)$\frac{25}{117}$
(c)$\frac{1}{50}$
(d)$\frac{3}{25}$
(e)None of these
Answer
Answer (as printed): A
Explanation
Let $S$ be the sample space and let $E$ be the event of selecting 2 boys and 1 girl. Then, $n(S) \quad=$ number of ways of selecting 3 students out of $25={ }^{25} C_{3}=\frac{25 \times 24 \times 23}{3 \times 2 \times 1}=2300$. And, $n(E)=\left({ }^{15} C_{2} \times{ }^{10} C_{1}\right)=\left(\frac{15 \times 14}{2 \times 1} \times 10\right)=1050$. $$\therefore \quad P(E)=\frac{n(E)}{n(S)}=\frac{1050}{2300}=\frac{21}{46} .$$
Explanation as extracted from the printed page; notation may be imperfect.