Question

An oil drilling company ventures into various locations, and its success or failure is independent from one location to another. Suppose the probability of a success at any specific location is 0.25.
(a) What is the probability that the driller drills at 10 locations and has 1 success?
(b) What is the probability that the driller drills at 10 locations and has at least 2 success?

Answer 1

Answer:

a) 18.77% probability that the driller drills at 10 locations and has 1 success

b) 75.60% probability that the driller drills at 10 locations and has at least 2 success

Step-by-step explanation:

For each drill, there are only two possible outcomes. Either it is a success, or it is not. The probability of a drill being a success is independent of other drills. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

10 locations

This means that $n = 10$

Suppose the probability of a success at any specific location is 0.25.

This means that $p = 0.25$

(a) What is the probability that the driller drills at 10 locations and has 1 success?

This is P(X = 1).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{10,1}.(0.25)^{1}.(0.75)^{9} = 0.1877$

18.77% probability that the driller drills at 10 locations and has 1 success

(b) What is the probability that the driller drills at 10 locations and has at least 2 success?

Either there are less than 2 success, or there are at least 2. The sum of the probabilities of these events is decimal 1. So

$P(X < 2) + P(X \geq 2) = 1$

We want $P(X \geq 2)$

So

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.25)^{0}.(0.75)^{10} = 0.0563$

$P(X = 1) = C_{10,1}.(0.25)^{1}.(0.75)^{9} = 0.1877$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.0563 + 0.1877 = 0.2440$

$P(X < 2) = P(X = 0) + P(X = 1) = 1 - 0.244 = 0.756$

75.60% probability that the driller drills at 10 locations and has at least 2 success