Answer:
I) 0.386 = 38.6% probability that exactly one six occurs.
II) 0.518 = 51.8% probability that at least one six occurs.
Step-by-step explanation:
For each throw, there are only two possible outcomes. Either it is a six, or it is not. The probability of a six being rolled is independent of other throws, 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.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
Thrown four times
This means that [tex]n = 4[/tex]
Probability of rolling a six:
Six sides, one of which is 6. So
[tex]p = \frac{1}{6} = 0.1667[/tex]
i) exactly one six occurs
This is [tex]P(X = 1)[/tex]. So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 1) = C_{4,1}.(0.1667)^{1}.(0.8333)^{3} = 0.386[/tex]
0.386 = 38.6% probability that exactly one six occurs.
ii) at least one six occurs
This is:
[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]
So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{4,0}.(0.1667)^{0}.(0.8333)^{4} = 0.482[/tex]
[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.482 = 0.518[/tex]
0.518 = 51.8% probability that at least one six occurs.
