[\dfrac{2}{91}] A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The pr

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A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is:

$ \dfrac{1}{22} $

$ \dfrac{3}{22} $

$ \dfrac{2}{91} $

$ \dfrac{2}{77} $

Explanation:

Let S be the sample space.

Then, $ n \left(S\right)$ = number of ways of drawing 3 balls out of 15

= ¹⁵C₃

=$ \dfrac{(15 \times 14 \times 13)}{(3 \times 2 \times 1)} $

= 455.

Let E = event of getting all the 3 red balls.

$\therefore n \left(E\right)$ = ⁵C₃ = ⁵C₂ =$ \dfrac{(5 \times 4)}{(2 \times 1)} $= 10.

$\therefore P\left(E\right)$ =$ \dfrac{n(E)}{n(S)} $=$ \dfrac{10}{455} $=$ \dfrac{2}{91} $.

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