Using expectation/variance algebra in normal distribution A shop sells apples and pears. The masses, in grams, of the apples may be assumed to have a A~$N(180, 12^2)$ distribution and the masses of the pears, in grams, may be assumed to have a P~$N(100, 10^2)$ distribution. Find the probability that the mass of a randomly chosen apple is more than double the mass of a randomly chosen pear.
Although this question appeared in the linear combination of normal variables section, I wasn't quite sure how to incorporate it into this particular problem. I thought that it could just simply be 
$P(A>2P)$, where 2P can be found by multiplying the mean, 100, by 2. So it would be 
$P(A>200)$. But I don't think this approach is correct. How should I solve this? The answer is 0.196.

Guide:
If assume independence, then we have $A-2P$ is normal distribution with mean $$\mathbb{E}[A-2P]=\mathbb{E}[A]-2\mathbb{E}[P]=180-2(100)$$
and 
$$Var[A-2P]=Var[A]+4Va[P]=12^2+4(10^2)$$
