Question

A computer’s random number generator produces random integers from 1 to 50. What is the probability that at least one of the first 4 generated numbers is in the 20s?

0.0016

0.1024

0.4096

0.5904

Answer 1

0.5904

Explanation:

For each number, there are only two possible outcomes. Either it is in the 20s, or it is not. The probability of a number being in the 20s is independent of other numbers. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

In which
`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

4 generated numbers:

This means that
`n = 4`

Probability of a number in the 20s:

From 1 to 50, there are 50 numbers. Of those(from 20 to 29), there are 10 in the 20s. So
`p = (10)/(50) = 0.2`

What is the probability that at least one of the first 4 generated numbers is in the 20s?

Either none is, or at least one is. The sum of the probabilities of these events is 1. So

`P(X = 0) + P(X \geq 1) = 1`

We want
`P(X \geq 1)`. So

`P(X \geq 1) = 1 - P(X = 0)`

In which

`P(X = 0) = C_(4,0).(0.2)^(0).(0.8)^(4) = 0.4096`

`P(X \geq 1) = 1 - P(X = 0) = 1 - 0.4096 = 0.5904`