Suppose we have a machine that makes tires. Every minute the machine tries to make exactly one tire. The probability of success depends on the tire. If it does not succeed, the machine will try again to produce that tire in the next minute. Assume the probability that the $i$ tire will be success is $p_i$. I am interested in the expected number of tires after $m$ minutes in a specific example.
Assume we wait $m=3$ Minutes and we have $p_1 = 0.5 = p_2$.So in total a maximum of 2 tires can be produced
So the probability that we have exactly
1) $0$ tires after $3$ minutes is $0.5^3 = 0.125$.
2) 1 tire after 3 minutes is $0.5^3 + 0.5^3 + 0.5^3 = 0.375$.
3) 2 tires after 3 minutes is $0.5^3 + 0.5^3 + 0.5^3 = 0.375$.
So the expected number should be $0.125*0 + 0.375*1 + 0.375*2 = 1.125$. But the solution says that the correct value is $1.375$.
Did I make a mistake or the solution?
The error would seem to be in your calculation; allegedly, you have exhausted all possibilities of what might happen (we make 0, 1, or 2 tires), yet the total probability doesn't sum to 1.
I think the error is this: your comment that "in total a maximum of 2 tires can be produced" leads me to believe that the machine only attempts to make one of each type. If it makes type 1 successfully in the first minute, and makes type 2 successfully in the second minute, then it should shut off for the third minute. This would change your calculation of the probability of 2 tires being made, which should instead be $0.5^{\color{red}{2}} + 0.5^3 + 0.5^3 = \color{red}{0.5}$, which would then yield the expected value found in the book.