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Probability question

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

Reply 1

TeeEm

Original post by Mathstat

by definition

F(t) </= P(T</=t)

Reply 2

Mathstat

Original post by TeeEm

by definition

F(t) </= P(T</=t)

So it's just 1-e^(-λτ)?

Reply 3

TeeEm

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

Reply 4

Mathstat

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?

Reply 5

TeeEm

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.

Reply 6

Mathstat

Original post by TeeEm

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)

Thanks for your time

Reply 7

ghostwalker

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:

P(T\leq t) = P(T=0) + P(0 < T \leq t)

Edit: For t > 0.

And you're told in the question

P(0<T\leq t\; |\; T \not= 0)

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

Can you take it from there?

(edited 8 years ago)

Reply 8

TeeEm

Original post by Mathstat

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:

P(T\leq t) = P(T=0) + P(0 < T \leq t)

And you're told in the question

P(0<T\leq t\; |\; T \not= 0)

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.

Reply 9

Mathstat

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:

P(T\leq t) = P(T=0) + P(0 < T \leq t)

Edit: For t > 0.

And you're told in the question

P(0<T\leq t\; |\; T \not= 0)

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)?

Thanks for your help

Reply 10

ghostwalker

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,

P(T \leq t) = P(T=0) + P(0 < T \leq t)

= P(T=0) + P(0 < T \leq t\; | \; T\not= 0)P(T\not=0)

= p + (1-e^{-\lambda t})(1-p)

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

Thanks for your help

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

P(0\leq T<t \cap T \not= 0) = P(0 < T < t)

= P(0 < T \leq t\; | \; T\not= 0)P(T\not=0)

= (1-e^{-\lambda t})(1-p)

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.