I am currently learning binomial distributions and stuck on this particular problem. I was wondering if you can help me solve the following probability question:
A car salesman estimates his chances of selling a car to be $10\%$ on the first meeting with a customer, and $75\%$ on the second. It is assumed that all sales are made independently. Suppose that during one week of work, the salesperson sees five customers for the first time. During the same week, the salesperson also sees four other customers for the second time. What is the probability that he will make exactly two sales to the nine customers he meets during that week?
I thought finding the intersection between the two probabilities multiplied by the binomial formula for $X˜B(9;0.075)$, where $(X=2)$ would work, and have tried multiple other approaches but unfortunately to no end. I am getting quite frustrated. Thank you for the help in advance!
Following the comment, using $X_1$ and $X_2$ for the number of sales to first- and second-timers respectively: $$P=P(X_1=2,X_2=0)+P(X_1=X_2=1)+P(X_1=0,X_2=2)=\binom52(0.1)^2(0.9)^3\binom40(0.25)^4+\binom51(0.1)(0.9)^4\binom41(0.75)(0.25)^3+\binom50(0.9)^5\binom42(0.75)^2(0.25)^2=897399/6400000=0.140\dots$$