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

To find the probability of having 1 success out of 10 trials with a probability of success of 0.25, use the binomial probability formula. The probability is approximately 20.51%. To find the probability of having at least 2 successes in 10 trials, sum the probabilities of having 2, 3, 4, ..., or 10 successes. The probability is approximately 5.94%.

Step-by-step explanation:

This is a geometric problem because you may have a number of failures before you have the one success you desire. To find the probability that the driller drills at 10 locations and has 1 success, we can use the binomial probability formula:

P(X=k) = C(n,k) * p^k * (1-p)^(n-k),

where X is the number of successes, n is the number of trials, k is the specific number of successes we want, and p is the probability of success.

In this case, we want to find the probability of having 1 success out of 10 trials, with a probability of success of 0.25:

P(X=1) = C(10,1) * 0.25^1 * (1-0.25)^(10-1)

P(X=1) = 10 * 0.25 * 0.75^9

P(X=1) = 0.20508, or approximately 20.51%.

To find the probability that the driller drills at 10 locations and has at least 2 successes, we need to find the sum of the probabilities of having 2, 3, 4, ..., or 10 successes:

P(X>=2) = P(X=2) + P(X=3) + ... + P(X=10)

P(X>=2) = C(10,2) * 0.25^2 * 0.75^8 + C(10,3) * 0.25^3 * 0.75^7 + ... + C(10,10) * 0.25^10 * 0.75^0

P(X>=2) ≈ 0.05936, or approximately 5.94%.

Answer 2

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) = (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