# Question #afb17

##### 1 Answer

#### Answer

#### Answer:

#### Explanation

#### Explanation:

#### Answer:

At a rate of 12 strikes per minute, the probability of no strikes in one minute is about

It would take 3 min 20 sec (i.e. 40 strikes) before the probability of the reservoir being struck is at least

#### Explanation:

We can calculate the relative amount of area of the reservoir compared to the area of the whole storm:

#(3xx2)/(25xx7)=6/175#

This is also the probability that any single lightning bolt hits the reservoir.

Let

#"Pr"(X=x)=(1-p)^(x-1)p#

#color(white)("Pr"(X=x))=(169/175)^(x-1)(6/175)#

and

#"Pr"(X <= x)=1-(1-p)^x#

#color(white)("Pr"(X <= x))=1-(169/175)^x#

for

Given that strikes occur 12 times a minute, we seek the probability that it takes more than 12 strikes

#"Pr"(X>12)=1-"Pr"(X<=12)#

#color(white)("Pr"(X>12))=1-[1-(169/175)^12]#

#color(white)("Pr"(X>12))=(169/175)^12#

#color(white)("Pr"(X>12))~~0.6579#

*This is consistent with the answer you got. However, I notice that if the rate of lightning strikes is increased to 15 per minute, we get #P(X>15)~~0.5926,# so the answer key may have used a rate of 15/min.*

For part 2, we wish to answer how many lightning strikes (i.e. how many intervals of 5 seconds) it will take before there's a 75% chance the reservoir has been struck.

#"Pr"(X<=x)=1-(169/175)^x=3/4#

#color(white)("Pr"(X<=x)=1-)(169/175)^x=1/4#

#=>x=log_(169//175)(1/4)#

#color(white)(=>x)=log(1/4)/log(169/175)~~39.74#

We round up to get 40 lightning strikes, which gives

#"40 strikes"/"1 strike/5 sec"="200 sec"="3 min 20 sec".#

*Again, this is presuming the strike rate is 12/min. At a rate of 15/min, the answer would be 2 min 40 sec.*

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