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You have a set of ten light bulbs - the lifetime of each of them being given by an exponential RV with mean 1000 hrs. Find the probability that....

(a) at least 7 of the bulbs function for 1500 or more hours?

(b) no light bulb lasts more than 2000 hours?

My attempts:

(a) $\int_0^{1500} \frac{1}{1000}*e^{\frac{-1}{1000}*x} \,dx$ for one light bulb, but I don't see how to apply this to 7 out of 10 light bulbs

(b) $\int_0^{2000} \frac{1}{1000}*e^{\frac{-1}{1000}*x} \,dx$? but this doesn't seem right since we want no light bulbs...

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To $a):$

You get $P(X\le 1500)=F(1500)=1-e^{-\frac{3}{2}}.$ So $P(X\ge 1500)=1-F(1500)=e^{-\frac{3}{2}},$ which is the probability that a bulb lasts $1500$ hours or more. Now you need to consider a binomial distribution $Y\in B(10,e^{-\frac{3}{2}})$ and get $P(Y\ge 7).$ That is:

$$P(Y\ge 7)=P(Y=7)+P(Y=8)+P(Y=9)+P(Y=10)\\=\binom{10}{7}(e^\frac{-3}{2})^7(1-(e^\frac{-3}{2}))^3+\binom{10}{8}(e^\frac{-3}{2})^8(1-(e^\frac{-3}{2}))^2\\+\binom{10}{9}(e^\frac{-3}{2})^9(1-(e^\frac{-3}{2}))+\binom{10}{10}(e^\frac{-3}{2})^{10}$$

To $b):$

In a complete similar way, you have $P(X\le 2000)=F(2000)=1-e^{-2}.$ So $P(X\ge 2000)=1-F(2000)=e^{-2}.$ Now consider the binomial distribution $Y\in B(10,e^{-2})$ and get $P(Y=0).$

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  • $\begingroup$ I see now! Awesome thank you!! $\endgroup$ – user161154 Jul 2 '14 at 0:29
  • $\begingroup$ @user161154 You need to use a binomial distribution $B(n,p)$ with probability $p(X=k)=\binom{n}{k}p^k(1-p)^{n-k}.$ $\endgroup$ – mfl Jul 2 '14 at 0:30
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Start by finding the prob that a bulb does not last 1500 hours. : p=$\int_0^{1500} \frac{1}{1000}*e^{\frac{-1}{1000}*x}$. Find $q=1-p.$ Prob that atleast 7 bulbs function function for 1500 or more= $\sum _{i=7 to 10}(^{10}C_{i})q^{i}p^{10-i}$

Prob that a bulb does not last more than 2000 hours, r= $\int_{0}^{2000} \frac{1}{1000}*e^{\frac{-1}{1000}*x}. $ Hence, prob that no light bulb lasts more than 2000 hours= $r^{10}$

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