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?

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.

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

And p is the probability of X happening.

10 locations

This means that

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

This means that

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

This is P(X = 1).

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

We want

So

In which

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

d

step-by-step explanation:

given

y + 3 = - 4(x - 2) ← distribute the parenthesis

y + 3 = - 4x + 8 ( subtract 3 from both sides )

y = - 4x + 5 → d