Question

Find the probability that none of the three bulbs in a traffic signal must be replaced during the first 2700 hours of operation if the probability that a bulb must be replaced is a random variable b with density f(b) = 6(0.25 – (b - 2.5)2) when 2 < b < 3 and 0 otherwise, where b is time measured in multiples of 1000 hours. Find the probability that none of the three bulbs in a traffic signal must be replaced during the first 2700 hours of operation if the probability that a bulb must be replaced is a random variable b with density f(b) = 6(0.25 – (b - 2.5)2) when 2 < b < 3 and 0 otherwise, where b is time measured in multiples of 1000 hours. @Mathematics

Answer 1

1-f(2.7)^3

Answer 2

there should be an integral in your solution

Answer 3

OK, it's a density... so integrate 0 to 2.7 f(b) to get the probability of failure for one bulb in the first 2700 hours?

Answer 4

2 to 2.7

Answer 5

yeah, i have an integral but is the range from 2 to 2.7. that is what i got. i don't understand how it should be from 0 to 2.7

Answer 6

Yep, I meant 2...

Answer 7

the answer i got from the integral is 183.184. is that correct?

Answer 8

no...it is a density...it must be <= 1

Answer 9

okay. heres what i did. I simplified the equation f(b) to be equal to -6b^2 +30b-36 and integrated that with the range from 2 to 2.7. Is that the right equation to use or am i doing it wrong

Answer 10

do u-substitution

Answer 11

I'll be back on later(need to go to my office)...if you haven't received a good answer I'll help you.

Answer 12

okay. thank you anyway for your help so far

Answer 13

After you integrate, you get -2b^3 + 5b^2- 36b evaluated between 2 and 2.7
this gives 0.784
so, back to the problem, prob of no failures is 1-0.784^3

Answer 14

P(no failures)=P(red doesn't fail and yellow doesn't fail and green doesn't fail)

if we assume independence

=P(red doesn't fail)P(yellow doesn't fail)P(green doesn't fail)
let \[p=\int\limits_{2.7}^{3}f(b)db\]

then P(red doesn't fail)P(yellow doesn't fail)P(green doesn't fail)
\[=p\times p\times p=p^3\]

or let \[q=\int\limits_{2}^{2.7}f(b)db\]

then the prob of not failing is 1-q

then P(red doesn't fail)P(yellow doesn't fail)P(green doesn't fail)
\[=(1-q)\times(1-q)\times(1-q)=(1-q)^3\]

=.010077696 in either case

Answer 15

what is the value for q though after integration? I keep getting big numbers despite u substitution

Answer 16

.216

Answer 17

is 1-q

Answer 18

I see I am out to lunch this morning. But at least I can answer how to evaluate the integral:

\[\int\limits_{2}^{2.7}-6b^2 +30b-36 db= -2b^3+15b^2-36b \mid_2 ^{2.7}=\]
\[-2(2.7)^3+15(2.7)^2-36(2.7) - (-2(2)^3 +15(2)^2-36(2))\]