Question

Suppose approximately 75% of all marketing personnel are extroverts, whereas about 65% of all computer programmers are introverts. (For each answer, enter a number. Round your answers to three decimal places.)

(b)

In a group of 4 computer programmers, what is the probability that none are introverts?

What is the probability that 2 or more are introverts?
What is the probability that all are introverts?

Answer 1

Solution :

Let X be a random variable which represents the number of introvert computer programmers in a group of 4 programmers.

Given that, 65% of all computer programmers are introverts.

Let's consider finding an introvert computer programmer as success.

Probability of success (p) = 0.65

Number of trials (n) = 4

Since, we have only two mutually exclusive outcomes (success and failure) for each trials, probability of success remains constant in each of the trials, number of trials are finite and outcomes are independent, therefore we can consider that X follows binomial distribution with parameters n = 4 and p = 0.65.

According to binomial probability law, the probability of occurrence of exactly x successes in n trials is given by,

Where, p is probability of success.

b(i) We have to find P(X = 0).

We have, n = 4 and p = 0.65

Using binomial probability law we get,

Hence, In a group of 4 computer programmers, the probability that none are introverts is 0.015.

b(ii) We have to find P(X ≥ 2)

We have, n = 4 and p = 0.65

P(X ≥ 2) = 1 - P(X < 2)

P(X ≥ 2) = 1 - [P(X = 0) + P(X = 1)]

Using binomial probability law we get,

Hence, the probability that 2 or more are introverts is 0.874.

b(iii) We have to find P(X = 4).

We have, n = 4 and p = 0.65

Using binomial probability law we get,

Hence, the probability that all are introverts is 0.179.

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