2005 Sept B6

Hello,

I'm stuck on (i) - the answer in the examiner's report makes sense but I used a different method and although I found the right answer in the end (yay) I also found something else that puzzles me..

Here's what the question asks:
Studies of the lifetimes of a certain type of electric light bulb have shown that the probability of failure, q0 , during the first day of use is 0.05 and after the first day of use the force of failure , x , is constant at 0.01.

(i) Calculate the probability that a light bulb will fail within the first 20 days.

I tried calculating the probability of failing within 20 days directly instead of doing 1-P(not failing within 20 days).

The following formula gives the right answer:
0.05 + 0.95*Integral[0 to 19] {exp(-0.01*t)*0.01} dt
where
0.05 = probability of failing during day 1
0.95 = probability of surviving during day 1
exp(-0.01*t) = probability of surviving from day 0 to day t
0.01 = instantaneous rate of failure at time t (which is constant..)
Using latex: \[ 0.05 + 0.95 \int_{0}^{19} e^{-0.01t}0.01 dt \]
However, my initial formula (which made more sense to me, intuitively) is wrong and I don't see why:
0.05 + 0.95*Integral[1 to 20] {exp(-0.01*t)*0.01} dt
where
0.05 = probability of failing during day 1
0.95 = probability of surviving during day 1
exp(-0.01*t) = probability of surviving from day 1 to day 1+t
0.01 = instantaneous rate of failure at time t (which is constant..)
Using latex: \[ 0.05 + 0.95 \int_{1}^{20} e^{-0.01t}0.01 dt \]
Ultimately, the only difference between the 2 formulas is the limits of the integral..
Going from 1 to 20 makes more sense to me since we are given the probability of failing on the first day (ie day 0).

If anyone has any idea why this is wrong please tell meee!

thanks
Anj

 

Integrating a function between \(a\) and \(b\) gives the area under the curve and the area under that curve isn't the same for intervals \([0,19] \) and \([1,20] \).
\[ -[e^{-0.01t}]_0^{19} =0.173 \\ -[e^{-0.01t}]_1^{20} =0.171\]

 

Thanks for the correction.

 

No one has any ideas? Any help would be appreciated.

 

Constant force means distribution of lifetime from time 1 onward is exponential. Since exponential distribution has memory-less property, remaining time will start from 0 again. So the limits should be from 0 to 19.

 

To fail within 20 days, it can either
1) fail on the first day (q0) or
2) survive the first day and fail during the next 19 days (p0 * 19q1).

nqx = integral from 0 to n of tpx mu(x+t) dt.

So in this case 19q1 = int from 0 to 19 of tp1 mu(1+t) dt = int from 0 to 19 of 0.01e^(-0.01t) dt.