The problem says: A bombing plane flies directly above a railroad track. Assume that if a large bomb falls within 40 feet of the track, the track will be sufficiently damaged so that traffic will be disrupted. Let X denote the perpendicular distance from the track that a bomb falls. Assume that
$f_x\left(x\right)=\frac{100-x}{5000}\:I_{\left(0,100\right)}\left(x\right)$
a) Find the probability that a large bomb will disrupt the traffic. b) If the plane can carry three large bombs and uses all three, what is the probability that traffic will be disrupted?
What I did for part a) was:
$\int _0^x\frac{100-x}{5000}\:=\frac{1}{\:5000}\int _0^x\left(100-x\right)dx\:=\frac{1}{\:5000}\int _0^{40}\left(100-x\right)dx\:=\frac{1}{\:5000}\left(100\left(40\right)-\frac{\left(40\right)^2}{2}\right)=F\left(40\right)-F\left(0\right)=0.64-0=0.64$
And for part b) By Bernoulli's distribution
$P\left(x=1\right)=\left(0.64\right)^1\left(1-0.64\right)^{3-1}=\left(0.64\right)\left(0.36\right)^2=0.083$
Thanks in advance if you can confirm my procedure. \end{document}
Part (a) looks OK.
Part (b) is incorrect. The probability of disruption with three bombs should not be lower than the probability of disruption with one bomb.
There are a couple of things going on here:
(1) You are calculating the probability for exactly one bomb causing disruption, when you should be thinking about at least one bomb.
(2) You have a scenario with multiple trials (i.e. binomial distribution) but you're plugging it into the formula for a Bernoulli distribution, which assumes a single trial. Because of this, your formula is missing a term.