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Two persons, A and B are each picking up one piece of candy from a basket consisting 100 Kit-Kats, 200 Almond Joys, 300 Whoppers and 400 Skittles. A likes Skittles; B likes Almond Joys. What is the probability that between the two of them, they will get one Skittles and one Almond Joy, and what is the probability that two of them will get the same candy, of whatever type?

The solutions are:

P(they get one Skittles and one Almond Joy) = 0.16

P(they get the same candy) = 0.30

I'm not sure how the numbers come up...

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up vote 2 down vote accepted

There are $1000$ candies, of which $400$ are Skittles (S) and $200$ Almond Jpys (AJ). We can analyze the problem in two ways: (i) Imagine the picks are consecutive or (ii) Imagine that the householder grabs simultaneously two pieces of candy.

The probability the first pick is skittles is $\dfrac{400}{1000}$. If the first pick was S, the probability the next is AJ is $\dfrac{200}{999}$. So the probability of S then AJ is $\dfrac{400}{1000}\cdot \dfrac{200}{999}$.

Similarly, the probability of AJ then S is $\dfrac{200}{1000}\cdot\dfrac{400}{999}$, the same. So our desired probability is $$2\dfrac{400}{1000}\cdot \dfrac{200}{999}.$$

We could also count. Imagine the candies all have ID numbers. There are $(1000)(999)$ equally likely ways of picking two candies, counting order. And there are $(400)(200)$ ways of picking S, then AJ. There are $(200)(400)$ ways to pick AJ, then $S$. So our probability is $\dfrac{(400)(200)+(200)(400)}{(1000)(999)}.$

For way (ii), there are $\dbinom{1000}{2}$ ways to pick a pair of candies. And there are $(400)(200)$ ways to end up with one S and one AJ. Divide.

For the second problem, again we could count order or not, and we could work directly with probabilities, or count.

We will do it by looking at the $(1000)(999)$ ways of picking two candies, in order.

There are $(100)(99)$ ways to pick two Kit-Kats, $(200)(199)$ to pick two Almond Joys, and so on. So our probability is $$\dfrac{(100)(99)+(200)(199)+(300)(299)+(400)(399)}{(1000)(999)}.$$

Or else one can say that there are $\dbinom{1000}{2}$ ways to pick two candies, order of picking not counting. There are $\dbinom{100}{2}$ ways that these two are Kit-Kats, $\dbinom{200}{2}$ ways they are Almond Joys, and so on. When we calculate this way, we end up he same answer, though it looks a little different.

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