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

Answer:

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

Step-by-step 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.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

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:

$p = \frac{4}{6} = 0.6667$

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.