Probability, when to know whether the order matters. A car salesman can make a sale to 65% of his male customers but to only 45% of his female customers. All of his sales are independent. On Monday morning, the car salesman has two make and one female customer. Find the probability that he makes exactly two sales.
My Solution:
Lets Call the two male customers Tom and Harry and the female customer Jane.
Case 1: Tom bought, Harry didn't and Jane bought = (0.65)(0.35)(0.45)=0.102375
Case 2: Harry bought, Tom didn't and Jane bought = (0.65)(0.35)(0.45)=0.102375
Case 3: Harry bought, Tom bought and Jane didn't = (0.65)(0.65)(0.55)=0.232375
Case 4: Tom bought, Harry bought and Jane didn't = (0.65)(0.65)(0.55)=0.232375
Case 5: Jane didn't, Harry bought and Tom bought = (0.55)(0.65)(0.65)=0.232375
Case 5: Jane didn't, Tom bought and Harry bought = (0.55)(0.65)(0.65)=0.232375
Case 6: Jane bought, Tom bought and Harry didn't = (0.45)(0.65)(0.35)=0.102375
Case 7: Jane bought, Tom didn't and harry bought = (0.45)(0.35)(0.65)=0.102375
by now, adding up all the cases will lead to a probability of more than 1 which doesn't make sense. The correct answer is 0.437125
 A: Since each person has two choices, buy / don't buy, there can only be a total of $2^3 = 8$ distinct events
Your case $3$ and $4$, eg are exactly the same event, both males buy, and the female doesn't. Whether Tom buys first or Harry or Jane buy first doesn't come into the picture at all.
If you eliminate such duplications, the probabilities will sum up to $1$, as they must
A word of caution. Suppose Jane buys, and one of the males buys. Don't fall into the trap of considering this as one of the $8$ possibilities. Tom buying, Harry not buying is distinct from Harry buying, Tom not buying.
A: Hint: Notice that
$$(.35 + .65 x)^2 (.55 + .45 x) = 0.067375\, +0.305375 x+0.437125 x^2+0.190125 x^3$$
The coefficient of $x^2$ matches your book's answer.  Maybe this is not a coincidence.
A: There is no reason in this scenario why the order should matter. None of the probabilities depend in any way on the sequence of events.
Your cases 1-3 cover all the ways in which the salesman makes exactly two sales, so you simply need to sum them:
$$P(\text{exactly 2 sales})=0.102375+0.102375+0.232375=0.437125$$
If the customers can only go in one at a time, this would make no difference provided all three still go in eventually.  It would make a difference of course if having to wait leads to a customer leaving without seeing the salesman.
It may be helpful to contrast your scenario with one in which the following also applies: if the salesman makes a sale to the first customer, he feels more confident, so makes a better sales pitch, and as a consequence his probability of a sale to each subsequent customer is increased by 10%.  Then the sequence of events would matter.
More generally, the order matters in a scenario where the probabilities of outcomes of a later event depend on the outcome of an earlier event. Take the case of drawing 2 balls in turn from a bag containing 3 red and 4 blue balls, the question being to find the probability (P) of obtaining one red and one blue ball.  With replacement, the probabilities on drawing the second ball are the same as those on drawing the first, but without replacement they differ.  So with replacement, we can just calculate:
$$P = 2(3/7)(4/7) = 24/49$$
Without replacement we need to calculate the two possible sequences separately and then sum them:
$$P = (3/7)(4/6)+(4/7)(3/6) = 4/7$$
