1. ## rackets

A racket comes in midsize and oversize. $60 \%$ of all customers at a certain store want the oversize version.

The store currently has seven rackets of each version. What is the probability that all of the next ten customers who want this racket can get the version they want from current stock?

Is it: $\binom{7}{6}(0.6)^{6}(0.4) \times \binom{7}{4}(0.4)^{4}(0.6)^{3}$?

2. Originally Posted by shilz222
A racket comes in midsize and oversize. $60 \%$ of all customers at a certain store want the oversize version.

The store currently has seven rackets of each version. What is the probability that all of the next ten customers who want this racket can get the version they want from current stock?

Is it: $\binom{7}{6}(0.6)^{6}(0.4) \times \binom{7}{4}(0.4)^{4}(0.6)^{3}$?
I don't think so. Let X be the random variable number of customers who want an oversize racket.

Then the probability that the next ten customers get the racket they want is $\Pr(3 \leq X \leq 7)$.

Note: If less than 3 customers want the oversize, then more than 7 want the midsize and someone's gonna miss out .....

Now note that X ~ Binomial(n = 10, p = 0.6).

Therefore $\Pr(3 \leq X \leq 7) = \sum_{r = 3}^{7} {10 \choose r} (0.6)^r 0.4^{10 - r} = ......$