Kitty Oil Co. has decided to drill for oil in 10 different locations; the cost o

Question

Kitty Oil Co. has decided to drill for oil in 10 different locations; the cost of drilling at each location is S10,000. (Total cost is then $100,000.) The prob- ability of finding oil in a given location is only , but if oil is found at a given location, then the amount of money the company will get selling oil (excluding the initial $10,000 drilling cost) from that location is an exponential random variable with mean S50,000. Let Y be the random variable that denotes the number of locations where oil is found, and let Z denote the total amount of money received from selling oil from all the locations. (a) Find 8[Z]. (b) Find P[Z> 100,0001 Y-1] and P(Z > 100,000! Y= 2]. (c) How would you find P[Z> 100,000]? Is P[Z> 100,000] >? 22

Explanation / Answer

It is not clear whether Z

is income or net income. No big deal, if we can handle one we can handle the other. We use the gross income interpretation.

Let Z1,Z2,…,Z10

be the amount of money made from digs 1,2,…,10. Then Z=Z1+Z2++Z10. By the linearity of expectation, we have E(Z)=E(Z1)++E(Z10)=10E(Z1)

.

To find E(Z1)

, note that Z1=0 with probability 1p, where p is the probability of finding oil if one digs, currently unreadable. And given that the well was successful, the expectation is 50000. Thus E(Z1)=(1p)(0)+(p)(50000)

.

For the probability that Z>10000

given Y=1, we just want the probability that an exponential with mean 50000 is greater than 10000

.

Remark: If we interpret Z

as net income, for the expectation question subtract 100000

.

For the probability question, find the probability that an exponential with mean 50000

is greater than 110000.

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