Ambiguous(?) probability question: when to combine probabilities? The following question comes from Mathematical Statistics with Applications (7th ed.) by Wackerly, Mendenhall, and Schaeffer:

A merchant stocks a certain perishable item. She knows that on any given day she will have a
demand for either two, three, or four of these items with probabilities $.1$, $.4$, and $.5$, respectively.
She buys the items for $\$1.00$ each and sells them for $\$1.20$ each. If any are left at the end of
the day, they represent a total loss. How many items should the merchant stock in order to
maximize her expected daily profit?

The answer provided in the book says that three items should be bought, but I don't understand the reasoning in the manual, highlighted in bold:

There are three scenarios:

*

*if she stocks two items, both will sell with probability $1$. So, her profit is $\$0.40$.


*if she stocks three items, two will sell with probability $.1$ (a loss of $.60$) and three
will sell with probability $.9$. Thus, her expected profit is $$(–.60).1 + .60(.9) = \$0.48$$


*if she stocks four items, two will sell with probability .1 (a loss of $1.60$), three will
sell with probability $.4$ (a loss of $.40$), and four will sell with probability $.5$ (a gain
of $.80$. Thus, her expected profit is $$(–1.60).1 + (–.40).4 + (.80).5 = \$0.08$$
So, to maximize her expected profit, stock three items.

This value of $0.9$ is simply the probability of buying three items ($0.4$) added to the probability of buying four items ($0.5$)... I don't understand how this is justified. My thinking is the probabilities of more items being sold than bought should be discounted, leaving an $\frac{0.4}{0.4+0.1}=80\%$ chance of selling all three of the bought items and a $\frac{0.1}{0.4+0.1}=20\%$ chance of selling only two of them. This would make the expected profit from buying three items equal to $$ (-0.6)(0.2)+(0.6)(0.8)=\$0.36 $$
whereby buying two items maximizes the daily profit. Wouldn't the answer provided in the manual be more appropriate for the question

A merchant stocks a certain perishable item. She knows that on any given day, she will sell at least three items with probability $0.9$ and four items with probability $0.5$.
She buys the items for $\$1.00$ each and sells them for $\$1.20$ each. If any are left at the end of
the day, they represent a total loss. How many items should the merchant stock in order to
maximize her expected daily profit?

 A: The $.1, .4, .5$ are the probabilities of demand for $2,3,4$ items, not the probabilities she buys this number of items.  The demand is the number she will sell if she has enough available.  If she buys $3$ items and the demand is $3$ she will sell $3$.  If she buys $3$ items and the demand is $4$ she will sell $3$, so if she buys $3$ she has $.9$ chance of selling them all.  Your rewording works the same way.
A: 

She knows that on any given day she will have a
demand for either two, three, or four of these items with probabilities $.1$, $.4$, and $.5$, respectively. She buys the items for $\$1.00$ each and sells them for $\$1.20$ each.

*

*if she stocks three items, three  will sell with probability $.9$. Thus, her expected profit is $$(–.60).1 + \boldsymbol{.60(.9)} = \$0.48$$


On a day that she buys 3 pieces, there are are exactly two possibilities, disjoint:

*

*she sells 2 pieces, because there is demand for 2 pieces that day

*she sells 3 pieces, because there is demand for either 3 or 4 pieces that day; this happens with probability $(0.4+0.5)$ due to mutual exclusivity; in this case, her cost price is $\$(3\times1)$ and her selling price is $\$(3\times1.2)$ so her profit is $\$0.60.$

Wouldn't the answer provided in the manual be more appropriate for the question

A merchant stocks a certain perishable item. She knows that on any given day, she will sell at least three items with probability $0.9$ and four items with probability $0.5.$


If she buys 2 pieces that day, then she will sell at least 3 pieces with probability $0.$
