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}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
In which
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)!}](https://tex.z-dn.net/?f=C_%7Bn%2Cx%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
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](https://tex.z-dn.net/?f=p%20%3D%20%5Cfrac%7B4%7D%7B6%7D%20%3D%200.6667)
10 normal six sided dice are thrown.
This means that ![n = 10](https://tex.z-dn.net/?f=n%20%3D%2010)
Find the probability of obtaining at least 8 failures.
This is:
![P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10)](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%208%29%20%3D%20P%28X%20%3D%208%29%20%2B%20P%28X%20%3D%209%29%20%2B%20P%28X%20%3D%2010%29)
So
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
![P(X = 8) = C_{10,8}.(0.6667)^{8}.(0.3333)^{2} = 0.1951](https://tex.z-dn.net/?f=P%28X%20%3D%208%29%20%3D%20C_%7B10%2C8%7D.%280.6667%29%5E%7B8%7D.%280.3333%29%5E%7B2%7D%20%3D%200.1951)
![P(X = 9) = C_{10,9}.(0.6667)^{9}.(0.3333)^{1} = 0.0867](https://tex.z-dn.net/?f=P%28X%20%3D%209%29%20%3D%20C_%7B10%2C9%7D.%280.6667%29%5E%7B9%7D.%280.3333%29%5E%7B1%7D%20%3D%200.0867)
![P(X = 10) = C_{10,10}.(0.6667)^{10}.(0.3333)^{0} = 0.0174](https://tex.z-dn.net/?f=P%28X%20%3D%2010%29%20%3D%20C_%7B10%2C10%7D.%280.6667%29%5E%7B10%7D.%280.3333%29%5E%7B0%7D%20%3D%200.0174)
Then
![P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) = 0.1951 + 0.0867 + 0.0174 = 0.2992](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%208%29%20%3D%20P%28X%20%3D%208%29%20%2B%20P%28X%20%3D%209%29%20%2B%20P%28X%20%3D%2010%29%20%3D%200.1951%20%2B%200.0867%20%2B%200.0174%20%3D%200.2992)
0.2992 = 29.92% probability of obtaining at least 8 failures.