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URGENT HELP!!!
In a recent poll, 18% of seniors said they will attend the upcoming prom. Twenty seniors chosen at random were asked if they plan to attend the prom. What is the probability that at most 5 seniors will attend the prom?
Question 3 options:

0.136

0.149

0.758

0.864

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Using the binomial distribution, it is found that there is a 0.864 = 86.4% probability that at most 5 seniors will attend the prom.

What is the binomial distribution formula?

The formula is:

$$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$$

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

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, we have that:

- 18% of seniors said they will attend the upcoming prom, hence p = 0.18.

- Twenty seniors chosen at random were asked if they plan to attend the prom, hence n = 20.

The probability that at most 5 seniors will attend the prom is given by:

$$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$$

In which:

$$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$$

$$P(X = 0) = C_{20,0}.(0.18)^{0}.(0.82)^{20} = 0.0189$$

$$P(X = 1) = C_{20,1}.(0.18)^{1}.(0.82)^{19} = 0.0829$$

$$P(X = 2) = C_{20,2}.(0.18)^{2}.(0.82)^{18} = 0.1730$$

$$P(X = 3) = C_{20,3}.(0.18)^{3}.(0.82)^{17} = 0.2278$$

$$P(X = 4) = C_{20,4}.(0.18)^{4}.(0.82)^{16} = 0.2125$$

$$P(X = 5) = C_{20,5}.(0.18)^{5}.(0.82)^{15} = 0.1493$$

$$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0189 + 0.0829 + 0.1730 + 0.2278 + 0.2125 + 0.1493 = 0.864$$

0.864 = 86.4% probability that at most 5 seniors will attend the prom.

More can be learned about the binomial distribution at

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