Question
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:
Answer: Option C
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Let S be the sample space.
Then, n(S) = number of ways of drawing 3 balls out of 15
= 15C3
= \(\frac{(15\times14\times13)}{(3\times2\times1)}\)
= 455.
Let E = event of getting all the 3 red balls.
So, n(E) = 5C3 = 5C2 = \(\frac{(4\times5)}{(2\times1)}\) = 10.
So, \(P(E) = \frac{n(E)}{n(S)}=\frac{10}{455}=\frac{2}{91}.\)
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