Two large coffees among six orders

Question

At a cafe, small coffees are ordered as a Poisson process at rate $9$ per hour and large coffees independently at rate $3$ per hour. In one hour, exactly $6$ coffees were ordered in total.

Given that, what is the probability that exactly $2$ of them are large?

Accepted Answer

Given the total number of coffees, each order was independently small or large, with probabilities set by the rate shares, so the large-count is binomial. The chance any order is large is p = \frac{\lambda{	ext{large}}}{\lambda{	ext{small}} + \lambda{	ext{large}}} = \frac{3}{9 + 3} = 0.25. So the number of large coffees among 6 is Binomial(6, 0.25). The probability of exactly 2: P(2) = \binom{6}{2}(0.25)^2(0.75)^4 = 15 	imes 0.0625 	imes 0.316406 \approx 0.297. About a 30\% chance. Takeaway. Fix the total, and the split among types is binomial with p = that type's rate over the summed ra…

Two large coffees among six orders

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