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Sagot :
Using the binomial distribution, it is found that there is a 0.8909 = 89.09% probability that both outcome Y and outcome Z occur at least once.
The trials are independent, hence the binomial distribution can be used.
What is the binomial distribution formula?
The formula is:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The parameters are:
- x is the number of successes.
- n is the number of trials.
- p is the probability of a success on a single trial.
In this problem:
- 5 trials, hence [tex]n = 5[/tex].
For the probability that outcome Y does not happen, we have that:
- 0.1 probability of X, 0.5 of Z, hence [tex]p = 0.6[/tex].
The probability is P(X = 5), that is, all five outcomes are either X or Z, then:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 5) = C_{5,5}.(0.6)^{5}.(0.4)^{0} = 0.0778[/tex]
For the probability that outcome Z does not happen, we have that:
- 0.1 probability of X, 0.4 of Y, hence [tex]p = 0.5[/tex].
The probability is P(X = 5), that is, all five outcomes are either X or Y, then:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 5) = C_{5,5}.(0.5)^{5}.(0.5)^{0} = 0.0313[/tex]
Then, the probability that at least one does not happen is:
[tex]p = 0.0778 + 0.0313 = 0.1091[/tex]
The probability that both happen is:
[tex]1 - p = 1 - 0.1091 = 0.8909[/tex]
0.8909 = 89.09% probability that both outcome Y and outcome Z occur at least once.
You can learn more about the binomial distribution at https://brainly.com/question/24863377
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