according to a survey, 15% of city workers take the bus to work. donatella randomly surveys 10 workers. what…

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.\np(k\text{ successes}) = _nc_kp^k(1 - p)^{n - k}\n_nc_k=\frac{n!}{(n - k)!k!}\n0.001\n0.002\n0.128\n0.899

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.\np(k\text{ successes}) = _nc_kp^k(1 - p)^{n - k}\n_nc_k=\frac{n!}{(n - k)!k!}\n0.001\n0.002\n0.128\n0.899

Answer

Explanation:

Step1: Identify values

$n = 10$ (number of trials), $k = 6$ (number of successes), $p=0.15$ (probability of success in a single - trial).

Step2: Calculate combination ${n}C{k}$

${n}C{k}=\frac{n!}{(n - k)!k!}=\frac{10!}{(10 - 6)!6!}=\frac{10!}{4!6!}=\frac{10\times9\times8\times7}{4\times3\times2\times1}=210$.

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