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A kinder egg may contain a prize of type $A$, of type $B$, or be empty. The probability of a prize $A$ is $5$ times larger than of prize $B$. A box contains 12 kinder eggs, from which 7 are known to be empty. Three eggs are drawn at random.

a.) Find the prob. that you find at least one prize $A$

b.) Given that you've found exactly one prize $A$, find the prob. that the other two were $B$'s.

My attempt:

Since an egg can either contain $A,B$ or nothing, we have

$$p_a +p_b +p_0 = 1$$

Incorporating the condition $p_a =5p_b$, this leads to:

$$p_0=1-6p_a$$ $$p_a=p_a$$ $$p_b = \frac{1}{5} p_a$$

Since we know that 7 eggs are empty, we can say (though I'm not sure about this bit)

$$p_0 = \frac{N_0}{N} = \frac{7}{12}.$$

This gives us expressions for each of the probabilities, using the above system of equations.

For part a.) we have $P(A_{\ge 1}) = 1-P(A_0)$, so all that remains is to find $P(A_0)$, where the notation is clear. This is binomial, since no A's is equivalent to all B's, all empty, or three ways of either 2 B's, one empty or 2 empty, one B:

$$P(A_0) = (p_0 +p_b)^3.$$

For part b.) it's a bit unclear to me. For three drawn eggs, the probability of one A (the condition given) results from one of twelve outcomes, in groups of four, that is the total probability of getting exactly one A is:

$$P(A_1) = 4(p_a p_0 ^2) + 4(2p_a p_b p_0) +4(p_a p_b ^2)$$

So now, the probability of getting two B's given one A is

$$P(B_2|A_1)= \frac{P(B_2 \cap A_1)}{P(A_1)}$$

But the probability of two B's and one A is

$$P(B_2 \cap A_1) = 3p_b ^2 p_a$$

So with this all that remains is to plug in the numbers.

I would appreciate someone checking my work because I have the worst intuition when it comes to classical/combinatorial probability...

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1 Answer 1

up vote 1 down vote accepted

We do have $p_a=5p_b$ and $p_a+p_b+\frac{7}{12}=1$, and we can now calculate $p_b$ and $p_a$. So we can, as you did, take the basic probabilities as known.

Your solution to (a) is correct, it is now just a matter of filling in the numbers.

For (b), you are aware that we need to find $\Pr(A_1)$, the probability of exactly $1$ prize A. This one is tricky. There are $3$ ways this can happen. (i) We have $2$ empty eggs, and the remaining prize is an A; (ii) We have $1$ empty egg, and precisely one of the two remaining eggs has an A: (iii) We have $0$ empty eggs and precisely one of the three eggs has an A. To complete the calculation, you will also need to use the probability of (iii), as your solution indicated.

We calculate the probability of (i). There are $\binom{12}{3}$ equally likely ways to choose $3$ eggs. There are $\binom{7}{2}\binom{5}{1}$ ways to choose two empty eggs and a non-empty one. Thus the probability we choose two empty and one non-empty is $\frac{\binom{7}{2}\binom{5}{1}}{\binom{12}{3}}$.

Given that we got $2$ empty and $1$ non-empty, the probability our non-empty has an A is $\frac{p_a}{p_a+p_b}$. This ratio is $\frac{5}{6}$. So the probability of (i) is $$\frac{\binom{7}{2}\binom{5}{1}}{\binom{12}{3}}\cdot \frac{5}{6}.$$

Now we need the probabilities of (ii) and of (ii). The basic arguments are the same as for (i). We leave these to you. If difficulties arise, please leave a comment.

Remark: The tricky thing about the calculations is that both the hypergeometric and the binomial are involved. We could bypass the binomial coefficients by imagining we are choosing eggs one at a time.

But we have to take account of the fact that, for example, choosing an empty egg on the first trial affects the probability of choosing an empty egg on the next trial.

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Thank you for your answer. But there's still something I don't understand - in the problem I have to calculate the conditional probability. Is the formula I used correct: $P(B_2|A_1)=\frac{P(A_1 \cap B_2)}{P(A_1)}$, and did I calculate the numerator bit of this correctly in my answer? –  DepeHb Jul 1 '13 at 16:55
    
The numerator was not calculated correctly. The numerator is the answer to (iii). We can calculate it using the basic technique we used for (i). More simply, the probability of picking $3$ eggs with prizes is $(5/12)(4/11)(3/10)$. Given that happened, the probability of $1$ A and $2$ B is $\binom{3}{1}(5/6)(1/6)^2$. That is because $\frac{p_a}{p_a+p_b}=\frac{5}{6}$. So now I have calculated your numerator, and (iii). All you need is (ii). Your conditional proability setup is fine. –  André Nicolas Jul 1 '13 at 17:06
    
Ok, I think I understand. By the way, can the law of total conditional probability be used here? What would the partition $(H_n)$ be? I'm referring to $P(A|X)=\sum P(A|X\cap H_i)P(H_i | X)$. Here $A=B_2$ and $X=A_1$ –  DepeHb Jul 1 '13 at 17:10
    
One can use streamlined formulas, though exactly the same calculations will need to be made. Minor exception, we do not need to compute $\binom{12}{3}$. Out of general principle, I try to make sure that for these basic calculations, students proceed from the ground. Even withhold some formulas from them. –  André Nicolas Jul 1 '13 at 17:21
    
I've carefully gone over your answer and I understand it, minus one small step. In (i), how did you get the probability that the non empty egg has A? It's not at all obvious to me why it should be $p_a /(p_a +p_b)$. What I usually do in these problems is draw tables, and in this situation, it looks like this : (A,0,0) (0,A,0) (0,0,A) (B,0,0) (0,B,0) (0,0,B) . The components represent the content of each egg. From this, I see 3 cases of interest (the first three). How do I get from here to finding the probability? –  DepeHb Jul 2 '13 at 15:16

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