Suppose approximately 80% of all marketing personnel are extroverts, whereas abo

Question

Suppose approximately 80% of all marketing personnel are extroverts, whereas about 70% of all computer programmers are introverts. (Round your answers to three decimal places)

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

(b) What is the probability that 5 or more are extroverts?

(c) What is the probability that all are extroverts?

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

(a) What is the probability that 3 or more are introverts?

(b) What is the probability that all are introverts?

Explanation / Answer

(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 = 0.8

We have to find P(X10)

P(X10) = 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*0.8^10*0.2^5 = 0.103182294

P(X=11) = 15C11*0.8^11*0.2^4 = 0.187604171

P(X=12) = 15C12*0.8^12*0.2^3 = 0.250138895

P(X=13) = 15C13*0.8^13*0.2^2 = 0.230897442

P(X=14) = 15C14*0.8^14*0.2^1 = 0.131941395

P(X=15) = 15C15*0.8^15*0.2^0 = 0.035184372

P(X10) = 0.93894857

Required probability = 0.93894857

(b) What is the probability that 5 or more are extroverts?

We are given

n = 15, p = 0.8

We have to find P(X5)

P(X5) = 1 – P(X4)

P(X5) = 1 - 0.00001246

P(X5) = 0.99998754

Required probability = 0.99998754

(c) What is the probability that all are extroverts?

We are given

n = 15, p = 0.8

We have to find P(X=15)

P(X=x) = nCx*p^x*(1 – p)^(n – x)

P(X=15) = 15C15*0.8^15*0.2^0 = 0.035184372

Required probability = 0.035184372

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

We are given n = 5, p = 0.7

We have to find P(X=0)

P(X=x) = nCx*p^x*(1 – p)^(n – x)

P(X=0) = 5C0*0.7^0*0.3^5 = 0.00243

Required probability = 0.00243

(a) What is the probability that 3 or more are introverts?

We are given n = 5, p = 0.7

We have to find P(X 3)

P(X 3) = 1 – P(X2) = 1 - 0.16308

P(X 3) = 0.83692

Required probability = 0.83692

(b) What is the probability that all are introverts?

We are given n = 5, p = 0.7

We have to find P(X = 5)

P(X=x) = nCx*p^x*(1 – p)^(n – x)

P(X=5) = 5C5*0.7^5*0.3^0 = 0.16807

Required probability = 0.16807

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.