calculate the probability of normal distribution variable with conditions

Question

There're 3 brothers who are saving money every month. The 1st brother saves \$160 monthly; the savings of the 2nd brother obey normal distribution with $E=\$280$ ($E$ = expected value), $\sigma = 30$ ($\sigma$ = standard deviation); the savings of the 3rd brother are 1.5 times bigger than that of the 2nd brother. The monthly savings are independent in different months.

The parents of the brothers decide to add \$100 each month when the 1st and the 2nd brother save more than the 3rd brother. What is the probability that in a given month the parents will add money?

Let "brother" = $B$, then I think I need to calculate $P(B_3<B_2+B_1) = \phi(B_3) - \phi(B_2+B_1)$. But $B_1$ doesn't obey normal distribution, nor we know that $B_3$ does. And even if they did should all variables be equal of its expected value for the calculations?

Answer 1

$B_1$ is a constant (equals 160 with probability 1)

Now:

$$P(B_1+B_2>B_3)=P(160+B_2>1.5 B_2)=P(0.5 B_2<160)=P(B_2<320)$$

and i think you can solve the rest yourself.

Answer 2

From the question, $B_{1}=160$, $B_{2}\sim\mathcal{N}(280,30^2)$ and $B_{3}=1.5B_{2}$. You are looking for $\mathbb{P}[B_{1}+B_{2}\geq B_{3}]=\mathbb{P}[160+B_{2}\geq 1.5B_{2}]=\mathbb{P}[B_{2}\leq320]$. Since $B_{2}\sim\mathcal{N}(280,30^{2})$, $\frac{B_{2}-280}{30}\sim\mathcal{N}(0,1)$. Hence $\mathcal{P}[B_{2}\leq320]=\mathcal{P}[\frac{B_{2}-280}{30}\leq\frac{4}{3}]=\Phi(\frac{4}{3})\approx0.91$, where $\Phi$ is cdf of standard normal distribution.