Question

Approximately 7.3% of residents in the Charleston area use public transportation on a regular basis. If 29 residents are randomly selected, what is the probability that exactly one of them use public transportation on a regular basis.Round your answer to at least three decimals

Answer 1

Given:

probability of success (p) = 7.3% or 0.073 in decimal form

sample size = 29

Find: the probability of x = 1 (uses public transportation on a regular basis)

Solution:

Based on the given information, the probability of success is only 7.3%. From this, we can infer that the probability of failure would be 92.7% or 0.927 in decimal form.

`100\%-7.3\%=92.7\%=0.927`

Since there are only two possible outcomes here, uses public transportation (success) and not using public transportation (failure), we are dealing with binomial probability.

The formula for this is:

`P(x=1)=_nC_x* p^x* q^(n-x)`

where n = sample size, x = the number of success, p = success probability, and q = failure probability.

In the word problem, n = 29, x = 1, p = 0.073, and q = 0.927. Let's plug this into the formula above.

`P(x=1)=_(29)C_1*0.073^1*0.927^(28)`

Then, solve.

`\begin{gathered} P(x=1)=29*0.073*0.1197381534 \\ P(x=1)=0.2534857 \\ P(x=1)\approx0.253 \end{gathered}`

Answer:

The probability that exactly one of them uses public transportation on a regular basis is approximately 0.253.