SOLUTION: Q1 About 9% of people are left-handed. Suppose 5 people are selected at random.18 (a) What is the probability that all are right-handed? (b) What is the probability that all are

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Question 1104723: Q1 About 9% of people are left-handed. Suppose 5 people are selected at random.18
(a) What is the probability that all are right-handed?
(b) What is the probability that all are left-handed?
(c) What is the probability that not all of the people are right-handed?

Answer by stanbon(75887) (Show Source):
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Q1 About 9% of people are left-handed. Suppose 5 people are selected at random.18
P(left-handed) = 0.09 ; P(right-handed) = 0.91
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(a) What is the probability that all are right-handed?
P(x = 5 are right handed) = 18C5*0.91^5*0.09^13 = binompdf(18,0.91,5) = 1.36x10^-10
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(b) What is the probability that all are left-handed? = 0.09^18 = 1.5x10^-19
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(c) What is the probability that not all of the people are right-handed?
P(x = 18 are right-handed) = 1- 0.91^18 = 0.8169
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Cheers,
Stan H.
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