Question

Assume that when adults with smartphones are randomly​ selected, 55​% use them in meetings or classes. If 7 adult smartphone users are randomly​ selected, find the probability that at least 2 of them use their smartphones in meetings or classes.

Answer 1

Given that,

n = 7, p = 0.55

Thus X follows a binomial distribution with parameters n = 7 and p = 0.55

X~Bin(7, 0.55 )

The probability mass function of X is given below :

P(X=x) = nCx*p^x*q^(n – x)

the probability that at least 2 of them use their smartphones in meetings or classes.

P(X≥2) = 1 – P(X<2)

= 1 – P(X<=1)

=1- [P(X=0) + P(X=1) ]

=1-[ 7C0*0.55^0*0.45^7 + 7C1*0.55^1*0.45^6 ]

=1-[ 0.003736695 + 0.031969498 ]

=1- 0.03571

= 0.9643

Required probability = 0.9643