1
$\begingroup$

I'm trying to solve the following problem:

The number of bombs required to achieve the disintegration is assumed to be geometrically distributed. In one series in which y bombs are available, x of these bombs failed. What is the probability that it will disintegrated by the end?


My understanding & solution:

We know that the memoryless property states that: $P(X >s+t|X >t)=P(X >s)$

Now I think that the probability is: $1 - P(X>(y-x) +x | X >x)= 1 - P(X>(y-x))$

Here I have set :

  • t = the number of bombs which failed => x
  • s = the number of bombs available => (y-x)

Is my reasoning correct? Sorry for the lack of proper wording but this is all new to me and I'm still trying to join the dots.

$\endgroup$
1
$\begingroup$

Yes, your reasoning is absolutely correct. Let N be the number of bombardments required to achieve the disintegration, then N ~ Geometric(p) . Now since Geometric Distribution has memoryless property, the fact that we know that x bombardments have failed is not going to make the chances of disintegration more(or less) in the upcoming bombardments. That is the Probability that now the disintegration will happen within the remaining (y-x) bombardments is just as likely as if we are starting a series of fresh (y-x) bombardments.

Hence the required Probability is

P(N=1) + P(N=2) + P(N=3) + ... + P(N=(y-x))

which can be easily computed as it is just a geometric series of (1-p).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.