[\dfrac{21}{46}] In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability t

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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:

$ \dfrac{21}{46} $

$ \dfrac{25}{117} $

$ \dfrac{1}{50} $

$ \dfrac{3}{25} $

Explanation:

Let S be the sample space and E be the event of selecting 1 girl and 2 boys.

Then, $ n \left(S\right)$ = Number ways of selecting 3 students out of 25

= ²⁵C₃ `

=$ \dfrac{(25 \times 24 \times 23)}{(3 \times 2 \times 1)} $

= 2300.

$ n \left(E\right)$ = (¹⁰C₁ x ¹⁵C₂)

=$ \left(10 \times\dfrac{(15 \times 14)}{(2 \times 1)} \right) $

= 1050.

$\therefore P\left(E\right)$ =$ \dfrac{n(E)}{n(S)} $=$ \dfrac{1050}{2300} $=$ \dfrac{21}{46} $.

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