# A storm measuring 25m xx 7m is over a reservoir measuring 3m xx 2m. The storm produces 12 lightning strikes per minute. What is the probability of the reservoir not being struck in one minute? I get 0.6579 but the textbook answer is 0.5926.

## How long would it take before the probability of the reservoir being struck is at least $\frac{3}{4}$?

Oct 28, 2017

At a rate of 12 strikes per minute, the probability of no strikes in one minute is about $0.6579 .$

It would take 3 min 20 sec (i.e. 40 strikes) before the probability of the reservoir being struck is at least $\frac{3}{4.}$

#### Explanation:

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

$\frac{3 \times 2}{25 \times 7} = \frac{6}{175}$

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

Let $X$ be the number of lightning strikes needed until one hits the reservoir. Then $X \text{ ~ GEO} \left(p = \frac{6}{175}\right) :$

$\text{Pr} \left(X = x\right) = {\left(1 - p\right)}^{x - 1} p$

$\textcolor{w h i t e}{\text{Pr} \left(X = x\right)} = {\left(\frac{169}{175}\right)}^{x - 1} \left(\frac{6}{175}\right)$

and

$\text{Pr} \left(X \le x\right) = 1 - {\left(1 - p\right)}^{x}$

$\textcolor{w h i t e}{\text{Pr} \left(X \le x\right)} = 1 - {\left(\frac{169}{175}\right)}^{x}$

for $x = 1 , 2 , 3 , e t c .$

Given that strikes occur 12 times a minute, we seek the probability that it takes more than 12 strikes $\left(\text{12 strikes"/"min"xx"1 min}\right)$ before one hits the reservoir:

$\text{Pr"(X>12)=1-"Pr} \left(X \le 12\right)$

$\textcolor{w h i t e}{\text{Pr} \left(X > 12\right)} = 1 - \left[1 - {\left(\frac{169}{175}\right)}^{12}\right]$

$\textcolor{w h i t e}{\text{Pr} \left(X > 12\right)} = {\left(\frac{169}{175}\right)}^{12}$

$\textcolor{w h i t e}{\text{Pr} \left(X > 12\right)} \approx 0.6579$

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 $P \left(X > 15\right) \approx 0.5926 ,$ 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.

$\text{Pr} \left(X \le x\right) = 1 - {\left(\frac{169}{175}\right)}^{x} = \frac{3}{4}$

$\textcolor{w h i t e}{\text{Pr} \left(X \le x\right) = 1 -} {\left(\frac{169}{175}\right)}^{x} = \frac{1}{4}$

$\implies x = {\log}_{169 / 175} \left(\frac{1}{4}\right)$
$\textcolor{w h i t e}{\implies x} = \log \frac{\frac{1}{4}}{\log} \left(\frac{169}{175}\right) \approx 39.74$

We round up to get 40 lightning strikes, which gives

$\text{40 strikes"/"1 strike/5 sec"="200 sec"="3 min 20 sec} .$

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