I'm having trouble with a supposedly easy exercise in elementary probability. We're given that the probability of a boy being left-handed is $0.12$, and $0.08$ for a girl. Then in a group of 25 kids, 10 male and 15 female, the exercise asks me to find the probability of $2$ boys and $1$ girl being left-handed.
I thought the probability would be
$${10 \choose 2}\times0.12^2\times0.88^8\times{15 \choose 1}\times0.08\times0.92^{14}\approx0.0870$$
but the solution my teacher provided us with says
$${10 \choose 2}\times0.12^2\times0.88^8+{15 \choose 1}\times0.08\times0.92^{14}\approx0.6065$$
At first I thought the plus sign was nonsense, but now I'm just confused about this. Surely the probability of exactly 2 boys and 1 girl being left-handed in such a large group can't be 60%, I think...
$P(AB) = P(A)\cdot P(B)$ if A and B are independent. Your A and B (boys being left-handed and girls being left-handed) are independent, so your answer is correct.
One way to think about is that for likely events A and B the adding method could give a probability greater than 1.