Question

Suppose approximately 80% of all marketing personnel are extroverts, whereas about 60% 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 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)

P(extroverts) = 0.80

X ~ Binomial (n,p)

Where n = 15, p = 0.80

Binomial probability distribution is

P(X) = ⁿC_x p^x (1-p)^n-x

i)

P( 10 or more extroverts) = P( X >= 10)

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

= ¹⁵C₁₀ 0.80¹⁰ 0.20⁵ +¹⁵C₁₁ 0.80¹¹ 0.20⁴ +¹⁵C₁₂ 0.80¹² 0.20³ +¹⁵C₁₃ 0.80¹³ 0.20² +

¹⁵C₁₄ 0.80¹⁴ 0.20 +¹⁵C₁₅ 0.80¹⁵ 0.20⁰

= 0.939

ii)

P( 5 or more extroverts) = 1 - P( 4 or less extroverts)

= 1 - P( X <= 4)

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

= 1 - [¹⁵C₀ 0.80⁰ 0.20¹⁵ +¹⁵C₁ 0.80¹ 0.20¹⁴ +¹⁵C₂ 0.80² 0.20¹³ +¹⁵C₃ 0.803 0.20¹² +

¹⁵C₄ 0.80⁴ 0.20¹¹ ]

= 1 - 0.00001

= 0.99999

= 1.000

iii)

P( All are extroverts) = P( X = 15)

= ¹⁵C₁₅ 0.80¹⁵ 0.20⁰

= 0.035

b)

P(introverts) = 0.60

X ~ Binomial (n,p)

Where n = 5, p = 0.60

Binomial probability distribution is

P(X) = ⁿC_x p^x (1-p)^n-x

i)

P( none are introvert) = P( X = 0)

= ⁵C₀ 0.60⁰ 0.40⁵

= 0.010

ii)

P( 3 or more introverts) = P( X >= 3)

= P( X = 3) + P( X = 4) + P( X = 5)

= ⁵C₃ 0.60³ 0.40² +⁵C₄ 0.60⁴ 0.40 +⁵C₅ 0.60⁵ 0.40⁰

= 0.683

iii)

P( All are introverts) = P( X = 5)

= ⁵C₅ 0.60⁵ 0.40⁰  

= 0.078