Question

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Answer 1

Using the binomial distribution, it is found that there is a 0.7447 = 74.47% probability that fewer than 7 of them show up.

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, the values of the parameters are given as follows:

p = 0.7, n = 8.

The probability that fewer than 7 of them show up is given by:

[tec]P(X < 7) = 1 - P(X \geq 7)[/tex]

In which:

$P(X \geq 7) = P(X = 7) + P(X = 8)$

Then:

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

$P(X = 7) = C_{8,7}.(0.7)^{7}.(0.3)^{1} = 0.1977$

$P(X = 8) = C_{8,8}.(0.7)^{8}.(0.3)^{0} = 0.0576$

Then:

$P(X \geq 7) = P(X = 7) + P(X = 8) = 0.1977 + 0.0576 = 0.2553$

[tec]P(X < 7) = 1 - P(X \geq 7) = 1 - 0.2553 = 0.7447[/tex]

0.7447 = 74.47% probability that fewer than 7 of them show up.

More can be learned about the binomial distribution at brainly.com/question/24863377

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