Binomial Distribution question (prob. of sinking the ship)

[Guest]

Five missiles are fired at a ship, each missile having a prob'lity of 3/4 of sinking it. What is the probability that the ship sunk?

 

[pka]

There is a probability of \(\displaystyle \left( {\frac{1}{4}} \right)^5\) that the ship will not sink!

 

[mcrae]

to set this up as a binomial distribution the question would ahve to ask what are the chances of it being hit with n number of rockets, in which case you would write


(5Cn)(.75)^n(.25)^(5-n)

if you really really really want to do it using binomial distribution, substitute 0 for n and do 1-p(0). or you can substitute 1, 2, 3, 4, and 5 for n and add those probabilities up

 

[Guest]

PKA I disagree with you

The probability should be 1023/1024...

 

[pka]

Re: PKA I disagree with you

The probability should be 1023/1024...

What? ? ? \(\displaystyle \L
1 - \left( {\frac{1}{4}} \right)^5 = \frac{{1023}}{{1024}}.\)

If p is the probability that S happens then P(not S) is 1-p.
If p is the probability that no ship sinks then 1-p is probability that at least one will sink. This is ideal example for using complement probability.

 

[galactus]

Pka is correct, of course: \(\displaystyle 1-(\frac{1}{4})^{5}=\frac{1023}{1024}\)=99.9% chnace of sinking.

 

[tkhunny]

I'm going to have to disagree just a bit. We need to know how the missles are fired.

If they are all fired together, the nice Binomial is good.

If they are fired in succession, without waiting for the results of each, it is an entirely different distribution, since we can't sink it twice - a truncated geometric? 3/4*[1 + 1/4 + (1/4)^2 + (1/4)^3 + (1/4)^4] But this produces the same result, 1023/1024. I guess I'm a little surprised by that.

If they are fired in succession WITH waiting for the result, then it is a simple Bernoulli experiment and the probability is 3/4. Since five were fired, the first four must have missed.

 

[pka]

I'm going to have to disagree just a bit. We need to know how the missiles are fired.... But this produces the same result, 1023/1024. I guess I'm a little surprised by that.

Oh well this is a slow Friday.
You should not be “a little surprised by that”.
Suppose the probability is p that the missile finds its target. We agree to fire up to K times, stopping if we hit or if we fire K missiles. What is the probability of sinking the ship?
[Let q=1-p.]

\(\displaystyle \begin{array}{c}
p + qp + pq^2 + \cdots + pq^{K - 1} \\
p\left( {1 + p + q^2 + \cdots + q^{K - 1} } \right) \\
p\frac{{1 - q^K }}{{1 - q}} \\
p\frac{{1 - q^K }}{p} \\
1 - q^K \\
\end{array}\)

 

[tkhunny]

You should not be “a little surprised by that”.

Clarification:

I was a little surprised by my a priori expectation of a slightly different result. The a posteriori expectation caused me to walk through what you have just demonstrated, proving the error of my ways.