Question

Approximately 47% of smartphones users upgrade phones every two years. sample of 7 students, what at is the probability that at most 5 students upgrade their phone every two years?

Answer 1

Solution

Given that ,

p = 0.47

1 - p = 0.53

n = 7

Using binomial probability formula ,

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

P(X 5) = P(X = 0) + P(X =1) + P(X = 2) + P(X = 3)+P(X = 4) +P(X = 5)

= ((7! /0 ! (7-0)!) * 0.47⁰ * (0.53)^7-0 + ((7! /1 ! (7-1)!) * 0.47¹ * (0.53)^7-1 + ((7! /2 ! (7-2)!) * 0.47² * (0.53)^7-2 + ((7! /3 ! (7-3)!) * 0.47³ * (0.53)^7-3 + ((7! /4 ! (7-4)!) * 0.47⁴ * (0.53)^7-4​​​​​​​ + ((7! /5 ! (7-5)!) * 0.47⁵ * (0.53)^7-5​​​​​​​

=  ((7! / 0! (7)!) * 0.47⁰ * (0.53)⁷ + ((7! /1! (6)!) * 0.47¹ * (0.53)⁶ + ((7! / 2! (5)!) * 0.47² * (0.53)⁵ +((7! / 3! (4)!) * 0.47³ * (0.53)⁴ + ((7! / 4! (3)!) * 0.47⁴ * (0.53)³ + ((7! / 5! (2)!) * 0.47⁵ * (0.53)²

= 0.0117 + 0.0729 + 0.194 + 0.2867 + 0.2543 + 0.1353

Probability = 0.9549