Question

10 normal six sided dice are thrown.Find the probability of obtaining at least 8 failuresif a success is 5 or 6.

Answer 1

0.2992 = 29.92% probability of obtaining at least 8 failures.

Explanation:

For each dice, there are only two possible outcomes. Either a failure is obtained, or a success is obtained. Trials are independent, which 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.

A success is 5 or 6.

A dice has 6 sides, numbered 1 to 6. Since a success is 5 or 6, the other 4 numbers are failures, and the probability of failure is:

`p = (4)/(6) = 0.6667`

10 normal six sided dice are thrown.

This means that
`n = 10`

Find the probability of obtaining at least 8 failures.

This is:

`P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10)`

So

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

`P(X = 8) = C_(10,8).(0.6667)^(8).(0.3333)^(2) = 0.1951`

`P(X = 9) = C_(10,9).(0.6667)^(9).(0.3333)^(1) = 0.0867`

`P(X = 10) = C_(10,10).(0.6667)^(10).(0.3333)^(0) = 0.0174`

Then

`P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) = 0.1951 + 0.0867 + 0.0174 = 0.2992`

0.2992 = 29.92% probability of obtaining at least 8 failures.