Question

(5.2.24) Assume that when adults with smartphones are randomly​ selected, ​58% use them in meetings or classes. If 11 adult smartphone users are randomly​ selected, find the probability that fewer than 4 of them use their smartphones in meetings or classes.
The probability is

Answer 1

P = 58 % = 0.58

n = 11

Evaluate P(x) if x = 0, 1, 2, 3.

$P(x = 0) = \frac{11!}{0!(11-0)!} \times 0.58^0 \times (1 – 0.58)^{11-0} = 1 \times 0.58^0 \times 0.42^{11} = 0.000071 \\ P(x = 1) = \frac{11!}{1!(11-1)!} \times 0.58^1 \times (1 – 0.58)^{11-1} = 11 \times 0.58^1 \times 0.42^{10} = 0.001089 \\ P(x = 2) = \frac{11!}{2!(11-2)!} \times 0.58^2 \times (1 – 0.58)^{11-2} = 55 \times 0.58^2 \times 0.42^{9} = 0.007524 \\ P(x = 3) = \frac{11!}{3!(11-3)!} \times 0.58^3 \times (1 – 0.58)^{11-3} = 165 \times 0.58^3 \times 0.42^{8} = 0.031172$

Use the additional rule for disjoint events:

$P(x < 4) = P(x=0) + P(x=1) + P(x=2) + P(x=3) \\ P(x<4) = 0.000071 + 0.001089 + 0.007524 + 0.031172 = 0.039856$

Answer: 0.039856