Question

The probability that a randomly selected person is left-handed is 0.1. Suppose we are to randomly select 10 people. Let X be the number of people selected out of the 10 that are left-handed. Find the following probabilities:

(a) P(X = 2) =

(b) P(X ≤ 2) =

(c) P(X > 2) =

Answer 1

Solution:
Given in the question
P(Selected person is left handed) = 0.1
Number of people selected(n) = 10
Solution(a)
we need to calculate probability that 2 out of 10 are left handed, here we will use binomial distribution,
P(X=n, N,p) = NCn*(p^n)*(1-p)^(N-n)
P(X=2) = 10C2*(0.1^2)*(1-0.1)^(10-2) = 45*0.01*0.4346721 = 0.1937
So there is 19.37% probability that 2 out of 10 are left handed.
Solution(b)
we need to calculate probability that less than or equal to 2 out of 10 are left handed, here we will use binomial distribution,
P(X=n, N,p) = NCn*(p^n)*(1-p)^(N-n)
P(X<=2) = P(X=0) + P(X=1) + P(X=2) = 10C0*(0.1^0)*(1-0.1)^(10-0)+ 10C1*(0.1^1)*(1-0.1)^(10-1)+ 10C2*(0.1^2)*(1-0.1)^(10-2) = 0.3488 + 0.3874 + 0.1937 = 0.9298
So there is 92.98% probability that less than or equal to 2 out of 10 are left handed.
Solution(c)
we need to calculate probability that greater than 2 out of 10 are left handed, here we will use binomial distribution,
P(X=n, N,p) = NCn*(p^n)*(1-p)^(N-n)
P(X>2) = 1- P(X<=2) = 1 - 0.9298 = 0.0702
So there is 7.02% probability that greater than 2 out of 10 are left handed.