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.
[tex]100\%-7.3\%=92.7\%=0.927[/tex]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:
[tex]P(x=1)=_nC_x\times p^x\times q^{n-x}[/tex]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.
[tex]P(x=1)=_{29}C_1\times0.073^1\times0.927^{28}[/tex]Then, solve.
[tex]\begin{gathered} P(x=1)=29\times0.073\times0.1197381534 \\ P(x=1)=0.2534857 \\ P(x=1)\approx0.253 \end{gathered}[/tex]Answer:
The probability that exactly one of them uses public transportation on a regular basis is approximately 0.253.
