Question

A company makes Citi bikes. 95% (i.e., 0.95) pass final inspection. Suppose that 5 bikes are randomly selected.
a. What is the probability that exactly 4 of these 5 sports bikes pass final inspection?
b. What is the probability that less than 3 of these 5 sports bikes pass final inspection?

Answer 1

Using the binomial distribution, it is found that the probabilities are given as follows:

a) 0.2036 = 20.36%.

b) 0.0011 = 0.11%.

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 parameters are given as follows:

p = 0.95, n = 5.

Item a:

The probability is P(X = 4), hence:

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

$P(X = 4) = C_{5,4}.(0.95)^{4}.(0.05)^{1} = 0.2036$

Item b:

The probability is:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2).

Then:

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

$P(X = 0) = C_{5,0}.(0.95)^{0}.(0.05)^{5} \approx 0$

$P(X = 1) = C_{5,1}.(0.95)^{1}.(0.05)^{4} \approx 0$

$P(X = 2) = C_{5,2}.(0.95)^{2}.(0.05)^{3} = 0.0011$

Then:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0 + 0 + 0.0011 = 0.0011.

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

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