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# Probability question watch

1. Hi, I am stuck on this question.

The life time T of a particular brand of light bulb is modelled as follows. There is a probability p of the light-bulb blowing immediately (so that T= 0); given that the light bulb does not blow immediately, the probability of it having lifetime τ or less is 1-e^(-λτ)

(i) Write down the cumulative distribution function F(t) of T

I am totally confused, especially as it just says write down. Any help would be appreciated
2. (Original post by Mathstat)
Hi, I am stuck on this question.

The life time T of a particular brand of light bulb is modelled as follows. There is a probability p of the light-bulb blowing immediately (so that T= 0); given that the light bulb does not blow immediately, the probability of it having lifetime τ or less is 1-e^(-λτ)

(i) Write down the cumulative distribution function F(t) of T

I am totally confused, especially as it just says write down. Any help would be appreciated
by definition

F(t) </= P(T</=t)
3. (Original post by TeeEm)
by definition

F(t) </= P(T</=t)
So it's just 1-e^(-λτ)?
4. (Original post by Mathstat)
So it's just 1-e^(-λτ)?
I have not done exponential distributions for a while but that is what I made it
5. (Original post by TeeEm)
I have not done exponential distributions for a while but that is what I made it
But where does the lightbulb exploding instantly with probability p come into it?
6. (Original post by Mathstat)
But where does the lightbulb exploding instantly with probability p come into it?
I do not what the question is asking, but I would imagine P is merely notation.

As far as your question is concerned it makes sense
the exponential distribution is defined between 0 and inf

F(0) = 0
F(inf) =1

and by differentiation you can find the PDF , mean and all sorts.
7. (Original post by TeeEm)
I do not what the question is asking, but I would imagine P is merely notation.

As far as your question is concerned it makes sense
the exponential distribution is defined between 0 and inf

F(0) = 0
F(inf) =1

and by differentiation you can find the PDF , mean and all sorts.
Is this not correct (this is working I just asked my friend for).

P(T<τ|T=/=0)= P(T<τ union T=/=0)/P(T=/=0)

The LHS = 1-e^(-λτ), so P(T<τ union T=/= 0) = P(T<τ)= p(1-e^(-λτ)

8. (Original post by Mathstat)
But where does the lightbulb exploding instantly with probability p come into it?
I beg to differ with TeeEm (WADR). Although the question says "write down", implying no working is required, I think you do need some in this case:

Edit: For t > 0.

And you're told in the question

Unusual question for A-level. Is it A-level?

Can you take it from there?
9. (Original post by Mathstat)
Is this not correct (this is working I just asked my friend for).

P(T<τ|T=/=0)= P(T<τ union T=/=0)/P(T=/=0)

The LHS = 1-e^(-λτ), so P(T<τ union T=/= 0) = P(T<τ)= p(1-e^(-λτ)

(Original post by ghostwalker)
I beg to differ with TeeEm (WADR). Although the question says "write down", implying no working is required, I think you do need some in this case:

And you're told in the question

Unusual question for A-level. Is it A-level?

Can you take it from there?
I will leave you in the capable hands of ghostwalker, as stats is not my forte ...
All the best.
10. (Original post by ghostwalker)
I beg to differ with TeeEm (WADR). Although the question says "write down", implying no working is required, I think you do need some in this case:

Edit: For t > 0.

And you're told in the question

Unusual question for A-level. Is it A-level?

Can you take it from there?
I am going into my second year of uni and this is on the holiday sheets.

So is the answer (1-p)(1-e^(-λτ)) ?

What is P(0<=T<τ union T=/=0)? Is it just 1-e^(-λτ)?

11. (Original post by Mathstat)
I am going into my second year of uni and this is on the holiday sheets.

So is the answer (1-p)(1-e^(-λτ)) ?
For t>0,

What is P(0<=T<τ union T=/=0)? Is it just 1-e^(-λτ)?

P(0<=T<τ union T=/=0). I think you mean intersect there, rather than union.

I've not been consistent with the < or <= at the upper end as you've used it slightly differently, but it's not important as long as there are no discontinuities, which there aren't for t > 0.

So you cdf will be as I previously wrote for t>0, and "p" for t=0
Edit: The case t=0 is actually covered by the formula - ho hum.
12. (Original post by ghostwalker)
For t>0,

For clarity, by the law of total probability that second line is really:

But those final terms come to 0, since T can't simultaneously be >0 and =0.

Hence as I originally wrote.
13. (Original post by ghostwalker)
For t>0,

P(0<=T<τ union T=/=0). I think you mean intersect there, rather than union.

I've not been consistent with the < or <= at the upper end as you've used it slightly differently, but it's not important as long as there are no discontinuities, which there aren't for t > 0.

So you cdf will be as I previously wrote for t>0, and "p" for t=0
Thanks you so much . All my friends were stuck on this as well.

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