SOLUTION: The probability that an American CEO can transact business in a foreign language is .20. Ten
American CEOs are chosen at random. (a) What is the probability that none can transact
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American CEOs are chosen at random. (a) What is the probability that none can transact
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Question 255362: The probability that an American CEO can transact business in a foreign language is .20. Ten
American CEOs are chosen at random. (a) What is the probability that none can transact business
in a foreign language? (b) That at least two can? (c) That all 10 can? (d) Sketch the probability
distribution and discuss its appearance. Found 2 solutions by drk, stanbon:Answer by drk(1908) (Show Source):
You can put this solution on YOUR website! P(speak foreign language) = .20
P(NOT speak foreign language) = .80
(a) What is the probability that none can transact business in a foreign language?
answer: we want zero "YES" and 10 "NO" so,
nCr*(no)^10*(yes)^0
or
10c0*.8^10*.2^0 ~ .1074
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(b) That at least two can?
answer: This means that 0 or 1 cannot. It is easier to do this in negation.
1 - p(0 not) + p(1 not) = 1 - 10c0(.8)^0*(.2)^10 + 10c1*(.8)^1*(.2)^9 ~ .99999
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(c) That all 10 can?
answer; we want all 10 yes, so,
10c10*(.2)^10 ~ 1.02 x 10^-7
(d) Sketch the probability distribution and discuss its appearance.
The sketch should be a normal curve with cum sum = 1.
You can put this solution on YOUR website! The probability that an American CEO can transact business in a foreign language is .20. Ten American CEOs are chosen at random.
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P(a CEO cannot transact business) = 0.80
It's a binomial problem with n = 10 ; p = 0.8 ; x varies
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(a) What is the probability that none can transact business
in a foreign language?
P(x = 0) = 0.8^10 = 0.1074
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(b) That at least two can?
P(x >=2) = 1 - binomcdf(10,0.2,1) = 0.6242
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(c) That all 10 can?
P(x=10) = 0.2^10 = 0.000000102
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Cheers,
Stan H.
(d) Sketch the probability
distribution and discuss its appearance.
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