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} $.