I think it is the fourth one
Answer:
Consider the number of children to be x.
So, 144 coins were divided amongst x children.
For x-3 children, each child would have 16 coins
So, (x-3)*16=144
Solving for x, x-3=144/16
or x-3=9
or x=9+3=12
Answer is 12
P(most favorable outcome) = 1 -(0.03 +0.16 -0.01) = 0.82
_____
"repair fails" includes the "infection and failure" case, as does "infection". By adding the probability of "repair fails" and "infection", we count the "infection and failure" case twice. So, we have to subtract the probability of "infection and failure" from the sum of "repaire fails" and "infection" in order to count each bad outcome only once.
The probability of a good outcome is the complement of the probability of a bad outcome.
