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Problem

There are $15$ identical bags of candy each containing $20$ yellow, $15$ red, $5$ blue and $10$ green candies. $15$ children are each given their own candy bag and each randomly picks $12$ candies from their own bags.

What is the probability that at least two of the kids will have at least one green candy?

This what I get so far:

Each bag contains $N=50$ candies out of which $k=10$ are green. Each child draws $n=12$ times without replacement. Considering the number of "successes", drawing a green candy, this is a Hypergeometric distribution with parameters $N,k,n$.

Therefore, the probability that a child draws at least one candy is

$$1-\frac{40\choose12}{50\choose 12}.$$

Now, I need to calculate the probabilty of two of this 15 kids will have at least one green candy. I'm stuck here.


I show my progress, after the help:

I calculate P(X>=)=1-P(X=0) = 0.9539 Then Y=# of kids with green candy.

P(y>=2)= 1- 15Cn0 p^0 (1-p0)^15 - 15Cn1 p^1 (1-p)^14 = 1-[(1)(1)(0.0461)^15 - (15 (0.9539) (0.0461)^14] = aprox 1

After all this I'm thinking:

Is that true? The probability of at leat two kids have at least one green candy could be 100%?

Thanks, for your help community.

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  • $\begingroup$ Not 100%, could be approx 100% ! Mind it, the expected # of candies with each child is 3. $\endgroup$ – true blue anil Oct 5 '15 at 3:26
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Ok, let P(a child has at least one green candy) = $p$, which you have computed.

Now apply binomial$(15,p)$

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