Determine the indicated probability for a binomial experiment with the given number of trials n and the given success probability p. n =8, p = 0.6, P(3 or fewer) Group of answer choices

Respuesta :

Answer:

[tex]P(X \leq 3) = 0.1738[/tex]

Step-by-step explanation:

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.

In this question, we have that:

[tex]n = 8, p = 0.6[/tex]

P(3 or fewer)

[tex]P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)[/tex]

In which

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 0) = C_{8,0}.(0.6)^{0}.(0.4)^{8} = 0.0007[/tex]

[tex]P(X = 1) = C_{8,1}.(0.6)^{1}.(0.4)^{7} = 0.0079[/tex]

[tex]P(X = 2) = C_{8,2}.(0.6)^{2}.(0.4)^{6} = 0.0413[/tex]

[tex]P(X = 3) = C_{8,0}.(0.6)^{3}.(0.4)^{5} = 0.1239[/tex]

[tex]P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0007 + 0.0079 + 0.0413 + 0.1239 = 0.1738[/tex]

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