In the evening, every child in a kindergarten is independently picked up by its parents. The mother will pick up the child with a probability of $40\%$ and the father will pick up the child with a probability of $60\%$. If the mother decides to pick up the child, there is a $15\%$ chance that she will be late. If the father decides to pick up the child, there is a $20\%$ chance that he will be late.
(a) What is the probability that a child will be picked up late from kindergarten?
(b) If a particular child is not picked up from kindergarten too late, what is the probability of its father picking it up?
(c) Given that two different children are picked up by their fathers, what is the probability that both children will be picked up late?
We define the events:
$M:$ Child is picked up by its mother.
$F:$ Child is picked up by its father.
$L:$ Child is picked up late.
(a) $P(L) = P(M) \cdot P(L | M) \cdot P(F) \cdot P(L | F) = 0.4 \cdot 0.15 + 0.6 \cdot 0.2 = 0.18.$
(b) $P(F | L^c) = \dfrac{P(F) \cdot P(L^c | F)}{1-P(L)} = 0.585.$
(c) Since each child is picked up independently from other children: $P(L | F)^2 = (0.6 \cdot 0.2)^2 = 0.0144.$
My book however gives $0.04$ as an answer for $c$. What's wrong with my solution?
You write that: "(c) Since each child is picked up independently from other children: $P(L|F)^2=(0.6⋅0.2)^2=0.0144$."
What's wrong: It is given that the $2$ children are picked up by their fathers. You do not need to multiply by $0.6$, the probability that a child is picked up by their father. So the answer is $0.2^2 = 0.04$.