Approximately 70% of all marketing personnel are extroverts, whereas about 60% o

Question

Approximately 70% 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 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?

Explanation / Answer

(A)
Probability that a marketing personal is extrovert, p = 0.7

Probability that 10 or more are extrovert,
P(X>=10) = P(X=10) + P(X=11) + P(X=12) + P(X=13) + P(X=14) + P(X=15)
P(X = 10) = 15C10 * 0.7^10 * 0.3^5 = 0.2061
P(X = 11) = 15C11 * 0.7^11 * 0.3^4 = 0.2186
P(X = 12) = 15C12 * 0.7^12 * 0.3^3 = 0.1700
P(X = 13) = 15C13 * 0.7^13 * 0.3^2 = 0.0916
P(X = 14) = 15C14 * 0.7^14 * 0.3^1 = 0.0305
P(X = 15) = 15C15 * 0.7^15 * 0.3^0 = 0.0048

P(X>=10) = 0.7216

The probability that 5 or more are extroverts?
P(X>=5) = 1 - P(X<=4)
P(X<=4) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)
P(X<=4) = 0.00067
P(X>=5) = 1 - 0.00067 = 0.9993

The probability that all are extroverts?
P(X = 15) = 15C15 * 0.7^15 * 0.3^0 = 0.0048

x P(X=x) 0 P(X=0) 0.00000001 1 P(X=1) 0.00000050 2 P(X=2) 0.00000820 3 P(X=3) 0.00008294 4 P(X=4) 0.00058058 5 P(X=5) 0.00298029 6 P(X=6) 0.01159000 7 P(X=7) 0.03477001 8 P(X=8) 0.08113003 9 P(X=9) 0.14723599 10 P(X=10) 0.20613038 11 P(X=11) 0.21862313 12 P(X=12) 0.17004021 13 P(X=13) 0.09156011 14 P(X=14) 0.03052004 15 P(X=15) 0.00474756

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