In a typical day, 61% of U.S. adults go online to get news. You randomly select five U.S. adults. Find the probability that the number of U.S. adults who say they go online to get news is
a) Exactly 2.
b)At least two.
c)More than two
Solution
Given that ,
p = 0.61
q = 1 - p = 1 - 0.61=0.39
n = 5
Using binomial probability formula ,
(A)P(X = x) = (n C x) * p x * (1 - p)n - x
P(X = 2) = (5 C 2) * 0.61 2 * (0.39)3
=0.2207
probability0.2207
(B)
P(X 2 ) = 1 - P( x <2)
= 1 - P(X = 0) - P(X = 1)
= 1 -(5 C 0) * 0.61 0 * (0.39)5-(5 C 1) * 0.61 1 * (0.39)4
=1-0.0796
probability=0.9204
(C)
P(X > 2) = 1 - P(X 2)
= 1 - P(X = 0) - P(X = 1) - P(X = 2)
= 1 - (5 C 0) * 0.61 0 * (0.39)5-(5 C 1) * 0.61 1 * (0.39)4 -(5 C 2) * 0.61 2 * (0.39)3
= 1 - 0.3003
= 0.6997
Probability = 0.6997
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