Question

What's the probability of getting exactly 1 five when we roll a fair die 10
times?

Answer 1

`P(X = 1) = C_(10,1).((1)/(6))^(1).(1-(1)/(6))^(9) = 0.323`, the correct answer is given by option c.

Explanation:

For each roll, there are only two possible outcomes. Either it is a five, or it is not. The probability of a roll being a five is independent of any other roll. This means that the binomial probability distribution is used 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.

The die has 6 sides:

One is five, so:

`p = (1)/(6)`

10 tosses:

This means that
`n = 10`

Probability of getting exactly one five.

This is P(X = 1). So

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

`P(X = 1) = C_(10,1).((1)/(6))^(1).(1-(1)/(6))^(9) = 0.323`

`P(X = 1) = C_(10,1).((1)/(6))^(1).(1-(1)/(6))^(9) = 0.323`, the correct answer is given by option c.