Question

according to a survey, 15% of city workers take the bus to work. donatella randomly surveys 10 workers. what is the probability that exactly 6 workers take the bus to work? round the answer to the nearest thousandth.

$p(k\text{ successes}) = _nc_kp^k(1 - p)^{n - k}$

$_nc_k=\frac{n!}{(n - k)!k!}$

0.001
0.002
0.128
0.899

Explanation:

Step1: Identify values of n, k, p

n = 10 (number of workers surveyed), k = 6 (number of workers taking bus), p = 0.15 (probability of a worker taking bus)

Step2: Calculate combination \(_{n}C_{k}\)

\(_{10}C_{6}=\frac{10!}{(10 - 6)!6!}=\frac{10!}{4!6!}=\frac{10\times9\times8\times7}{4\times3\times2\times1}=210\)

Step3: Calculate \((1 - p)^{n - k}\)

\(1-p=1 - 0.15 = 0.85\), \(n - k=10 - 6 = 4\), so \((1 - p)^{n - k}=0.85^{4}=0.52200625\)

Step4: Calculate \(p^{k}\)

\(p^{k}=0.15^{6}=0.000011390625\)

Step5: Calculate probability \(P(k)\)

\(P(6)=_{10}C_{6}\times p^{6}\times(1 - p)^{4}=210\times0.000011390625\times0.52200625\approx0.001\)

Answer:

0.001