Question

Suppose approximately 75% of all marketing personnel are extroverts, whereas about 50% of all computer programmers are introverts (Round your answers to three decimal places.) (a) At a meeting of 15 marketing personnel, what is the probablity that 10 or more are extroverts? What is the probability that 5 or more are extroverts? What is the probability that all are extroverts? (b) In a group of 5 computer programmers, what is the probability that none are introverts? What is the probability that 3 or more are introverts? What is the probability that all are introverts?

Answer 1

(a)

At a meeting of 15 marketing personnel, what is the probability that 10 or more are extroverts?

Solution:

We are given

n = 15, p = 0.75

We have to find P(X≥10)

P(X≥10) = P(X=10) + P(X=11) + P(X=12) + P(X=13) + P(X=14) + P(X=15)

P(X=x) = nCx*p^x*(1 – p)^(n – x)

P(X=10) = 15C10*0.75^10*0.25^5 = 0.165146

P(X=11) = 15C11*0.75^11*0.25^4 = 0.225199

P(X=12) = 15C12*0.75^12*0.25^3 = 0.225199

P(X=13) = 15C13*0.75^13*0.25^2 = 0.155907

P(X=14) = 15C14*0.75^14*0.25^1 = 0.066817

P(X=15) = 15C15*0.75^15*0.25^0 = 0.013363

P(X≥10) = 0.851632

Required probability = 0.852

What is the probability that 5 or more are extroverts?

We are given

n = 15, p = 0.75

We have to find P(X≥5)

P(X≥5) = 1 – P(X≤4)

P(X≥5) = 1 - 0.000115

P(X≥5) = 0.999885

Required probability = 1.000

What is the probability that all are extroverts?

We are given

n = 15, p = 0.75

We have to find P(X=15)

P(X=x) = nCx*p^x*(1 – p)^(n – x)

P(X=15) = 15C15*0.75^15*0.25^0 = 0.013363

Required probability = 0.013

(b)

In a group of 5 computer programmers, what is the probability that none are introverts?

We are given n = 5, p = 0.5

We have to find P(X=0)

P(X=x) = nCx*p^x*(1 – p)^(n – x)

P(X=0) = 5C0*0.5^0*0.5^5 = 0.03125

Required probability =0.031

What is the probability that 3 or more are introverts?

We are given n = 5, p = 0.5

We have to find P(X ≥ 3)

P(X ≥ 3) = 1 – P(X≤2) = 1 – 0.5

P(X ≥ 3) = 0.5

Required probability = 0.500

What is the probability that all are introverts?

We are given n = 5, p = 0.5

We have to find P(X = 5)

P(X=x) = nCx*p^x*(1 – p)^(n – x)

P(X=5) = 5C5*0.5^5*0.5^0 = 0.03125

Required probability = 0.031

(Probabilities are calculated by using Binomial table or excel.)