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?


What is the probability that 5 or more are extroverts?


What is the probability that all are extroverts?


(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

Please note ⁿC_x = n! / [(n-x)!*x!]

Some important probabilities for faster calculations:

(a) ⁿC₀ = 1, eg ⁵C₀ = 1, ⁹⁹C₀ = 1

(b) ⁿC_n = 1, eg ¹⁰C₁₀ = 1, ⁷⁵C₇₅ = 1

(c) ⁿC₁ = n, , eg ⁵C₁ = 5, ⁹⁹C₁ = 99

Binomial Probability = ⁿC_x * (p)^x * (q)^n-x, where n = number of trials and x is the number of successes.

Also sum of probabilities from 0 till n = 1, i.e P(0) + P(1) + P(2) +.......+P(n) = 1

_____________________________________________________________________________

(a) Here n = 15,p = 80% = 0.8, q = 1 - p = 0.2

(i) P(10 or more are extroverts) = P(X 10) = P(10) + P(11) + P(12) + P(13) + P(14) + P(15)

P(X = 10) = ¹⁵C₁₀ * (0.8)¹⁰ * (0.2)^{15-10 = 5} = 0.1032

P(X = 11) = ¹⁵C₁₁ * (0.8)¹¹ * (0.2)^{15-11 = 4} = 0.1876

P(X = 12) = ¹⁵C₁₂ * (0.8)¹² * (0.2)^{15-12 = 3} = 0.2501

P(X = 13) = ¹⁵C₁₃ * (0.8)¹³ * (0.2)^{15-13 = 2} = 0.2309

P(X = 14) = ¹⁵C₁₄ * (0.8)¹⁴ * (0.2)^{15-14 = 1} = 0.1319

P(X = 15) = ¹⁵C₁₅ * (0.8)¹⁵ * (0.2)^{15-15 = 0} = 0.0352

Therefore P(X 10) = 0.1032 + 0.1876 + 0.2501 + 0.2309 + 0.1319 + 0.0352 = 0.9389 0.939

____________________________________________________________________

(ii) P(5 or more extroverts) = P(X 5) = P(5) + P(6) + P(7) + P(8) + P(9) + P(10) + P(11) + P(12) + P(13) + P(14) + P(15)

Since P(1) + P(2) + P(3) + ...............+ P(14) + P(15) = 1

Therefore P(X 5) = 1 - [P(0) + P(1) + P(2) + P(3) + P(4)]

P(X = 0) = ¹⁵C₀ * (0.8)⁰ * (0.2)^{15-0 = 15} =

P(X = 1) = ¹⁵C₁ * (0.8)¹ * (0.2)^{15-1 = 14} = 0.0000

P(X = 2) = ¹⁵C₂ * (0.8)² * (0.2)^{15-2 = 13} = 0.0000

P(X = 3) = ¹⁵C₃ * (0.8)³ * (0.2)^{15-3 = 12} = 0.0000

P(X = 4) = ¹⁵C₄ * (0.8)⁴ * (0.2)^{15-4 = 11} = 0.0000

[P(0) + P(1) + P(2) + P(3) + P(4) = 0 + 0 + 0 + 0 + 0 = 0.0000

Therefore P(X 5) = 1 - (0.0000) = 1.000

__________________________________________________________________

(iii) P(all are extroverts) = P(X = 15)

P(X = 15) = ¹⁵C₁₅ * (0.8)¹⁵ * (0.2)^{15-15 = 0} = 0.035

________________________________________________________________________

(b) Here n = 4, p = 70% = 0.7, q = 1 - p = 0.3

(i) P(None are introverts = P(X = 0)

P(X = 0) = ⁴C₀ * (0.7)⁰ * (0.3)^{4-0 = 4} = 0.008

____________________________________________________________________

(ii) P(2 or more are introverts) = P( X 2) = P(2) + P(3) + P(4)

P(X = 2) = ⁴C₂ * (0.7)² * (0.3)^{4-2 = 2} = 0.2646

P(X = 3) = ⁴C₃ * (0.7)³ * (0.3)^{4-3 = 1} = 0.4116

P(X = 4) = ⁴C₄ * (0.7)⁴ * (0.3)^{4-4 = 0} = 0.2401

Therefore P( X 2) = 0.2646 + 0.4116 + 0.2401 = 0.9163    0.916

_________________________________________________________________________

(iii) P(All are introverts) = P(X = 4)

P(X = 4) = ⁴C₄ * (0.7)⁴ * (0.3)^{4-4 = 0} = 0.240

_________________________________________________________________________