# Probability exercise without combinatorics

Three kids go to by some icecream. There are 10 parfumes.

What is the probability that they pick-up $1, 2$ or $3$ different parfume ?

The answer should be $0.01, 0.27$ and $0.72$ but I can't figure out how to find directly the result for $2$.

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I guess you should state that the kids choose independently and that each flavor is chosen with equal probability. –  Fabian Jan 12 '12 at 19:32

It's the probability that:

"the first two kids chose the same and the third differently, or, the first two kids chose differently and the third chose a parfume chosen before"

So, it's $$\textstyle P({\text{first two }\atop\text{chose same} }) \cdot P({\text{third chose different given}\atop\text{the first two chose same}} )+P({\text{first two}\atop\text{chose different}})\cdot P({\text{third chose one chosen before }\atop\text { given the first two chose diff.}})$$ The probability that the first two kids chose the same parfume is $1\over10$. In this case, the probability that the third kid chose differently is $9\over 10$.

The probability that the first two kids chose differently is $9\over 10$. In this case, the probability that the third kid chooses one of the two parfumes chosen by the first two is $2\over 10$.

So, the probability that exactly 2 parfumes were chosen is $${1\over10}\cdot{9\over 10}+{9\over10}\cdot{2\over 10}= {27\over100}.$$

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After the first kid got his, if the next kid randomly chooses from the 10, what's the probability that they're the same? So what's the probability they're different?

Now, given those two possible outcomes and their probabilities, the third kid chooses his randomly from the 10. If both of the first two were the same, then what's the probability the third kid also chose the same? Multiply that by the probability that both two were the same to start with. What's the probability that the third kid chose differently. Multiply that with the probability the first two were the same.

Now consider the second possible outcome: two different. What's the probability the third kid choose one of those two? What's the probability he chooses something different? Multiply those into the probability of the 2 different outcome.

Now we have 4 outcomes, but 2 of them are identical (2 same, 1 different or (2 different, 1 same). So we can add those two probabilities together.

Is that what you mean by "directly"?

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Yes. For instance for 1, the first kid can choose among the 10 parfumes, the second and the third have to choose the same. So the probability is 10*1*1/1000 = 0.01. For 3, the formula is 10*9*8/1000 (the first can choose the parfume he wants, the second all the others parfumes except the first one and so on. –  Nielsou Hacken-Bergen Jan 12 '12 at 19:24
I'm just unable to do it for the number 2 –  Nielsou Hacken-Bergen Jan 12 '12 at 19:27

Call the kids A, B, and C, or something more interesting. They are lined up in that order at the ice-cream stand. They can end up with exactly $2$ different flavours in $3$ different ways, which I will abbreviate as XYY, YXY, and YYX, hoping the meaning is clear.

What is the probability of the pattern XYY? Whatever flavour X is picked by A, the probability that B picks a different flavour is $\frac{9}{10}$, and then C has probability $\frac{1}{10}$ of picking the same flavour as B did. Thus the probability of the pattern XYY is $\frac{9}{10}\cdot\frac{1}{10}=\frac{9}{100}$.

What is the probability of the pattern YXY? Whatever flavour Y is picked by A, the probability B picks a different flavour X is $\frac{9}{10}$, and then the probability C picks the same flavour as A did is $\frac{1}{10}$. We conclude that the probability of the pattern YXY is $\frac{9}{100}$.

In a very similar way, we can show that the probability of the pattern YYX is $\frac{9}{100}$.

Add up. We get $\frac{27}{100}$.

Another way: This one may be forbidden by the title ( "without combinatorics"). Suppose that A, B, C choose at random. Record their choices, in order. There are $(10)(10)(10)$ possible records, all equally likely.

How many of these records have exactly $2$ flavours? The minority flavour can be chosen in $10$ ways. For each of these ways, the majority flavour can be chosen in $9$ ways. And once these have been chosen, the position of the minority flavour (that is, who got it) can be chosen in $3$ ways, for a total of $(10)(9)(3)$ ways. So our probability is $270/1000$.

Another way: The "indirect way" that you alluded to is very efficient. Simply add together the probabilities of exactly one flavour and exactly $3$ flavours ($0.01+0.72=0.73$) and subtract this from $1$.

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