Learn / Library / Statistics

Buys before sells

StatisticsMedium

Buy orders arrive as a Poisson process with rate $2$ per minute, and sell orders as an independent Poisson process with rate $1$ per minute. What is the probability that $2$ buy orders arrive before $2$ sell orders do?

Checking an answer needs JavaScript. The hint and the solution open without it.

Hint

Each new order is a buy with a fixed probability, independently of the earlier ones.

Solution

Each arriving order is a buy with probability $\frac{2}{3}$ , independently of the rest.
The buys win the race with the sequences buy-buy, buy-sell-buy and sell-buy-buy.
Their probability is $(\frac{2}{3})^{2} + 2 \times (\frac{2}{3})^{2} \times (\frac{1}{3}) = \frac{4}{9} + \frac{8}{27}$ , which equals $\frac{20}{27}$ .

Answer: $\frac{20}{27}$

Next question Back to the library