Question

A company is considering drilling four oil wells. The probability of success for each well is 0.40, independent of the results for any other well. The cost of each well is $200,000. Each well that is successful will be worth $600,000.

a) What is the probability that one or more wells will be successful?

b) What is the expected number of successes?

c) What is the expected gain?

d) What will be the gain if only one well is successful?

e) Considering all possible results, what is the probability of a loss rather than a gain?

f) What is the standard deviation of the number of successes?

Answer 1

Given problem is a binomial distribution problem, given data are as follows

n = 4, p = 0.4 q = 1-p = 1 - 0.4 = 0.6

the probabilities in binomial distribution can be found by using the formula P(X=x) = ⁿC_x p^x q^n-x

Probability of no well successful = P(X = 0) = ⁴C₀ (0.4)⁰(0.6)⁴ = 0.1296

Probability of one well successful = P(X = 1) = ⁴C₁ (0.4)¹(0.6)³ = 0.3456

Probability of one well successful = P(X = 2) = ⁴C₂ (0.4)²(0.6)² = 0.3456

Probability of one well successful = P(X = 1) = ⁴C₃ (0.4)³(0.6)¹ = 0.1536

Probability of one well successful = P(X = 1) = ⁴C₄ (0.4)⁴(0.6)⁰ = 0.0256

Probability distribution table

X	0	1	2	3	4
P(X)	0.1296	0.3456	0.3456	0.1536	0.0256

a) the probability that one or more wells will be successful P(X>=1) = P(X=1)+P(X=2)+P(X=3)+P(X=4) = 1 - P(X=0) = 1-0.1296 P(X>=1) = 0.8704

b) the expected number of successes = = 0*.1296+ 0.3456*1 + 0.3456*2 + 0.1536*3 + 0.0256*4 = 1.6

c) the expected gain = expected success - expenditure = 1.6*$600,000 - 4*$200,000 = 960,000-800,000 = $160,000

d) the gain if only one well is successful

if only one well is successful then worth = $600,000

expenditure = 4*$200,000 = $800,000

If only one well is successful then gain = $600,000 - $800,000 = -$200,000

it is loss of $200,000

e) probability of loss = 1 - probability of gain

gain will happen if two or more wells were successful, hence probability of gain = P(X>1) = 1 - P(X<=1) = 1 - 0.1296-0.3456

probability of gain = 0.5248

probability of loss = 1 - probability of gain = 1 - 0.5248 = 0.4752

f) Standard deviation = ?

Variance of Successes = E(X²) - (E(X))²

E(X²) = = 0²*.1296+ 0.3456*1² + 0.3456*2² + 0.1536*3² + 0.0256*4² = 3.52

Variance of Successes = 3.52 - 1.6² = 0.96

std. dev = square root (variance ) = 0.96^0.5 = 0.979