Question

The probability that a randomly selected individual in a certain community has made an online purchase is 0.48 . Suppose that a sample of 10 people from the community is selected, what is the probability that at most 3 of them has made an online purchase?

Write only a number as your answer. Round to 2 decimal places (for example 0.24). Do not write as a percentage.

Answer 1

Solution

Given that ,

p = 0.48

1 - p = 0.52

n = 10

x 3

Using binomial probability formula ,

P(X = x) = ((n! / x! (n - x)!) * p^x * (1 - p)^{n - x}

P(X 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

P(X 3 ) = ((10! /0 ! (10-0)!) * 0.48⁰ * (0.52)^10-0 + ((10! /1 ! (10-1)!) * 0.48¹ * (0.52)^10-1 + ((10! /2 ! (10-2)!) * 0.48² * (0.52)^10-2+ ((10! /3 ! (10-3)!) * 0.48³ * (0.52)^10-3

= ((10! /0 ! (10)!) * 0.48⁰ * (0.52)¹⁰ + ((10! /1 ! (9)!) * 0.48¹ * (0.52)⁹ + ((10! /2 ! (8)!) * 0.48² * (0.52)⁸ + ((10! /3 ! (7)!) * 0.48³ * (0.52)⁷

= 0.0014 + 0.0133 + 0.0554 + 0.1364

= 0.2067

= 0.21

The probability that at most 3 of them has made an online purchase is 0.21