Question on Probability involving binomial distribution

[sankalpmittal]

Homework Statement

P is the probability that a person aged x years will die in a year. Find the probability that out of 5 men A,B,C,D and E, each of x years, A will die in the year and be the first to die.

Homework Equations

The Attempt at a Solution

I fixed A in the first place with probability P. Then I made cases of no one dying, one dying, and so on till all dying, which got me : P x ( ⁴C₀P⁰(1-P)⁴ + ⁴C₁P¹(1-P)³ + ⁴C₂P²(1-P)²+ ⁴C₃P³(1-P)¹+ ⁴C₄P⁴(1-P)⁰ ) which got me P as the answer, which was wrong.

 

[Merlin3189]

What is the chance that he is the first to die?

And then, what other condition has to be met?

 

[wabbit]

Edit : sorry duplicate reply, didn't see first post

 

[sankalpmittal]

Only that he has to die first and has to die in the year, which is why I fixed P outside the bracket. Could you guide me as to what I am doing wrong, or not getting?

 

[Merlin3189]

1 - "Only that he has to die first and has to die in the year," The difficulty with this I think, is that these two events are not independent. If he dies this year, I think that increases his chance of being first.
So I was looking to express the result as the product of two independent events.

2- Is your result = P(he dies this year) * P( that 0,1,2,3 or 4 people die this year) ? That doesn't quite make sense to me.

3 - Since I said P(He is first to die) * P( He dies this year ) is not valid, because not independent, then I need
P(He is first to die) * P( Something that is independent of the order of dying, but ensures he dies this year )

 

[sankalpmittal]

But how to calculate that with the given value. The question is only asking for one year so another approach I thought that could be is that A dies this year while the others don't because there is no other variable or information to actually calculate if more than one are dying then in which order, or is there?

 

[Ray Vickson]

Homework Statement

P is the probability that a person aged x years will die in a year. Find the probability that out of 5 men A,B,C,D and E, each of x years, A will die in the year and be the first to die.

Homework Equations

The Attempt at a Solution

I fixed A in the first place with probability P. Then I made cases of no one dying, one dying, and so on till all dying, which got me : P x ( ⁴C₀P⁰(1-P)⁴ + ⁴C₁P¹(1-P)³ + ⁴C₂P²(1-P)²+ ⁴C₃P³(1-P)¹+ ⁴C₄P⁴(1-P)⁰ ) which got me P as the answer, which was wrong.

The binomial distribution only governs how many will die, not WHEN they will die. Let N be the number of others (not A) who die, for N = 0,1,2,3,4. The probability that Mr. A and n others die is $p \times C(4,n) p^n (1-p)^{4-n}$; basically, that is what you wrote above. However, it says nothing about whether Mr. A is first, and you need to address that issue as well.

Given that n+1 die (Mr. A + n others) you then want the probability that A is first. Can you see a rather obvious answer to that? Can you see how to put all this together?

 

[Merlin3189]

" that A dies this year while the others don't"
Is not good enough, because he can be the first to die and have one or more others die as well this year.

Ray has the same idea as me, though I think I have a shorter route.
We want P( A is first) * P( At least somebody dies this year)
and there is an easier way of getting the latter, than ΣP(A and n others die, for n= 0 to 4),
because P(at least somebody dies this year) = 1 - P(nobody dies this year)

 

[sankalpmittal]

Okay, so will this work?
Probability that Mr.A and n others will die is P x C(4,n)Pⁿ(1-P)^4-n
Now there will be (n+1)! ways of these deaths and out of these, n! ways will be favourable so we divide P x C(4,n)Pⁿ(1-P)^4-n by (n+1)?
Will this be right?

 

[Ray Vickson]

Okay, so will this work?
Probability that Mr.A and n others will die is P x C(4,n)Pⁿ(1-P)^4-n
Now there will be (n+1)! ways of these deaths and out of these, n! ways will be favourable so we divide P x C(4,n)Pⁿ(1-P)^4-n by (n+1)?
Will this be right?

Yes, and then you have to sum for n from 0 to 4. Believe it or not, the final result is simple, because the sum can be done in closed-form.

 

[sankalpmittal]

I am unable to do the summation. Can you give advice on how to Σ(C(4,n)Pⁿ(1-P)^4-n)/(n+1)

 

[Ray Vickson]

I am unable to do the summation. Can you give advice on how to Σ(C(4,n)Pⁿ(1-P)^4-n)/(n+1)

The summation has only five terms, and aside from the p and (1-p) , all coefficients and powers are just numbers, so you can write out the whole thing explicitly on one line. Or, you can just leave it in symbolic summation form, but you should specify the limits of summation (n = 0 to n = 4); for example, you can write Σ_{n=0..4} or something similar.

I won't show you how to do the sum, because I want you to submit work which is your own.

 

[sankalpmittal]

Thanks people. I was able to sum it yielding the correct answer.

 

[Ray Vickson]

Thanks people. I was able to sum it yielding the correct answer.

Just as a matter of interest: what was your final result?

 

[sankalpmittal]

It was (1-(1-P)⁵)/5

 

[Ray Vickson]

It was (1-(1-P)⁵)/5

Right. Did you get it be simplifying the sum? That is one way to do it, but another way is to note that for each of the 5 people, the probabilities P_j = P{person j dies and is first} are all equal for j = 1,2 ,...,5, and their sum is the probability that somebody (i.e., at least one person) dies. That probability is 1 - (1-p)^5, so each person's P_j is 1/5 of that.

Of course, you can generalize it to n people, giving the corresponding result [1-(1-p)^n]/n.

 

[sankalpmittal]

Yep. Thanks! The second method is shorter! :)

 

[haruspex]

Yep. Thanks! The second method is shorter! :)

This is Merlin's method in post #8.