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If a bag contains 3 black balls and 8 white balls, and 5 balls are picked out in total, what's the probability of taking all the black balls?

I tried

$$ \frac{3\times 2\times 1}{11\times 10\times 9}\times 3! $$

Because I thought that there are $3!$ different orders the black balls can be picked in, and on each choosing the probability of picking black is one of the factors of the numerator over one of the factors of the denominator.

But this is wrong, I don't think it accounts for choosing balls after having picked a white one.

What's the correct way to approach this problem.

  • 1
    Look up the hypergeometric distribution. – 2017-01-16

2 Answers 2

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You have $3 \text{ Black Balls}$ and $8\text{ White Balls}$.

So,total you have $3+8=11\text{ Balls}$.

You pick $5\text{ Balls}$.You pick out all the black balls out.So,out of the $5$ balls you pick,$3$ are black and the remaining $2$ are white.

You can pick $5$ balls out of $11$ total balls in $11\choose5$ ways.

Also,you can pick $3$ black and $2$ white balls out of $3$ black and $8$ white balls respectively in $3\choose3$ $\times$ $8\choose2$ ways.

So the required probability is

$$P=\frac{{3\choose3}\times{8\choose2}}{11\choose5}$$.

Hope this helps!!

1

The number of ways to pick all the black balls is $\binom{3}{3}\cdot\binom{8}{5-3}=28$.

The total number of ways to pick $5$ balls is $\binom{3+8}{5}=462$.

The probability is therefore $\frac{28}{462}\approx6\%$.