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Suppose there are a total of 15 balls in a box. Out of these 15, 5 are of red color and others are green. 11 balls are chosen from the box at random. All balls are equally likely.

What is the probability that out of these 11, at least 3 balls are red colored?

My approach to solve this problem:

Let random var X = Number of red balls.
X takes values {0,1,2,3,4,5}
We have to find P(X>=3)

P(X>=3) = P(X=3) + P(X=4) + P(X=5)

Each ball has a probability 1/15.

Number of ways to choose 11 balls out of 15 balls is (15) C (11).

How do I proceed further?

  • 2
    **Hint**: Have a look to the **hypergeometric distribution**, if it is without replacement. – 2017-01-15

1 Answers 1

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If you take $11$ from those $15$ at the same time it's like choosing $4$ from $15$, and those $4$ have $\{0,1,2\}$ red balls. Take $\binom{15}{4}$ as the total. You're looking for $\binom{5}{0}\binom{10}{4} + \binom{5}{1}\binom{10}{3} + \binom{5}{2}\binom{10}{2}$. Then

$$ \textrm{P}(A) = \frac{\binom{5}{0}\binom{10}{4} + \binom{5}{1}\binom{10}{3} + \binom{5}{2}\binom{10}{2}}{\binom{15}{4}} = \frac{12}{13}. $$

If you want the complement, then the probability should be $\frac{1}{13}$.

Note: This is the simplest approach, which I assume you need since you didn't list restrictions.

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    Nice, although in this case there is no real difference to going after the stated question ($11$ balls) directly: $$p(A) = \frac{\binom{5}{3}\binom{10}{8} + \binom{5}{4}\binom{10}{7} + \binom{5}{5}\binom{10}{6}}{\binom{15}{11}} = \frac{12}{13}$$ – 2017-01-15
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    Indeed, but the coefficients are much easier to compute. I thought about it because calculating $\binom{5}{0}$ is free, for example. – 2017-01-15
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    @Joffan Also, nice use of infinity in your profile. 8 ) – 2017-01-15