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:

1²¹/₄₆
2²⁵/₁₁₇
3¹/₅₀
4³/₂₅

Answer:- 1

Explanation:-

Solution:

Let S be the sample space and E be the event of selecting 1 girl and 2 boys.
Then, n(S) = Number ways of selecting 3 students out of 25.

\begin{aligned} =^{25} C_{3} \\ = \frac{(25 \times 24 \times 23)}{(3 \times 2 \times 1)} \\ = 2300. \\ n (E) = (^{10} C_{1} \times^{15} C_{2}) \\ = [10 \times \frac{(15 \times 14)}{(2 \times 1)}] \\ = 1050. \\ ∴ P (E) = \frac{n (E)}{n (S)} = \frac{1050}{2300} = \frac{21}{46} . \end{aligned}

$\eqalign{ & = {}^{25}{C_3} \cr & = \frac{{\left( {25 \times 24 \times 23} \right)}}{{\left( {3 \times 2 \times 1} \right)}} \cr & = 2300. \cr & n\left( E \right) = \left( {{}^{10}{C_1} \times {}^{15}{C_2}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left[ {10 \times \frac{{\left( {15 \times 14} \right)}}{{\left( {2 \times 1} \right)}}} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1050. \cr & \therefore P\left( E \right) = \frac{{n\left( E \right)}}{{n\left( S \right)}} = \frac{{1050}}{{2300}} = \frac{{21}}{{46}}. \cr}$