Let X be a binomial random variable with p = 0.7
a) For n = 3, find P(X=1)
b) For n = 5, find P(X≤3)
Solution
Given that ,
p = 0.7
1 - p = 1 - 0.7 = 0.3
Using binomial probability formula ,
P(X = x) = ((n! / x! (n - x)!) * p^{x} * (1 - p)^{n - x}
(a)
n = 3
x = 1
P(X = 1) = ((3! / 1! (2)!) * 0.7^{1} * (0.3)^{2}
= 0.189
Probability = 0.189
b)
n = 5
P(x 3) = P(X = 0) + P(x = 1) + P(x = 2) + P(x = 3)
= ((5! / 0! (5)!) * 0.7^{0} * (0.3)^{5} + ((5! / 1! (4)!) * 0.7^{1} * (0.3)^{4} + ((5! / 2! (3)!) * 0.7^{2} * (0.3)^{3} + ((5! / 3! (2)!) * 0.7^{3} * (0.3)^{2}
= 0.4718
P(x 3) = 0.4718
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