Question

Suppose approximately 80% 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 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)

here this is binomial with parameter n=15 and p=0.8

probability that 10 or more are extroverts :

P(X>=10)=1-P(X<=9)=

1-∑x=0x-1   (nCx)px(q)(n-x) =

0.939

probability that 5 or more are extroverts :

P(X>=5)=1-P(X<=4)=

1-∑x=0x-1   (nCx)px(q)(n-x) =

1.000

probability that all are extroverts =0.8¹⁵ =0.035

b)

here this is binomial with parameter n=5 and p=0.5

probability that none are introverts =(1-0.5)⁵ =0.031

probability that 3 or more are introverts :

P(X>=3)=1-P(X<=2)=

1-∑x=0x-1   (nCx)px(q)(n-x) =

0.500

probability that all are introverts =0.5⁵=0.031