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An appliance comes in two colors, white and brown, which are in equal demand. A certain dealer in these appliances has three of each color in stock, although this is not known to the customers. Customers arrive and independently order these appliances. How do I find the probability that

a) the third white is ordered by the fifth customer?

b) all of the whites are ordered before any of the browns are ordered?

c) all of the whites are ordered before all of the browns are ordered?

Is there a more efficient method to solve these problems rather than counting?

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up vote 1 down vote accepted

A small amount of counting is used in (a). For (c), there is an easy no counting solution.

(a) We want the probability that the first four customers order a total of $2$ white and $2$ brown and the $5$th customer orders white. The probability of $2$ white and $2$ brown in $4$ orders is $\dbinom{4}{2}\left(\dfrac{1}{2}\right)^4$. Given that this has happened, the probability the $5$th order is white is $\dfrac{1}{2}$. For the required probability, multiply. We get $$\binom{4}{2}\left(\frac{1}{2}\right)^5.$$

(b) This is just the probability of three heads in a row when tossing a fair coin.

(c) This can be computed in various ways. One reasonable approach is by recycling. Let $x$ be the probability all the whites are ordered before any brown. Let $y$ be the probability the $3$rd white is the $4$th item ordered. And let $z$ be the probability that the $3$rd white is the $5$th item ordered. We want $x+y+z$. We already computed $x$ in part (b), and $z$ in part (a). It remains to compute $y$, which can be done in a way quite similar to the solution of (a).

But there is a trivial way of finding the answer! By symmetry, "we run out of whites before we run out of browns" is just as likely as "we run out of browns before we run out of whites." Thus the required probability is $\dfrac{1}{2}$.

Remark: The $\binom{4}{2}\left(\frac{1}{2}\right)^4$ that formed part of the answer to (a) did use counting. We need to know how many $4$-letter words there are that have $2$ W and $2$ B.

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