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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}

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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.

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  • $\begingroup$ Ohh I see. So with the formula N! / R! (N-R)! PR QN-R would be enough? $\endgroup$ – Josmar Muñoz May 19 '18 at 20:14
  • $\begingroup$ @JosmarMuñoz Missing some brackets and ^ symbols there, but otherwise that looks correct as the binomial formula for "out of N bombs, exactly R cause disruption, given that each bomb has probability P of doing so and Q = 1-P". $\endgroup$ – Geoffrey Brent May 19 '18 at 21:48
  • $\begingroup$ Awesome, thank you very much! $\endgroup$ – Josmar Muñoz May 20 '18 at 19:38
  • $\begingroup$ Sorry, I'm trying with "out of 3 bombs, exactly 3 or 1 cause disruption, given that each bomb has probability 0.64 of doing so and Q = 0.36" and I'm still getting a lower probability than with one single bomb, do you happen to know why is that? $\endgroup$ – Josmar Muñoz May 20 '18 at 21:56
  • $\begingroup$ And using the formula (N/(R!(N-R)!))*(P^R)(Q^(N-R)) $\endgroup$ – Josmar Muñoz May 20 '18 at 21:58

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