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_k} p^k (1 - p)^{n - k} \\)
\\( {_nc_k} = \frac{n!}{(n - k)! cdot k!} \\)
\\( \circ \\ 0.001 \\)
\\( \circ \\ 0.002 \\)
\\( \circ \\ 0.128 \\)
\\( \circ \\ 0.899 \\)

Explanation:

Step1: Define variables

Let $n=10$ (total workers), $k=6$ (successes), $p=0.15$ (bus probability)

Step2: Calculate combination $_{n}C_{k}$

$$ _{10}C_{6} = \frac{10!}{(10-6)! \cdot 6!} = \frac{10!}{4! \cdot 6!} = 210 $$

Step3: Compute $p^k$ term

$$ p^k = 0.15^6 = 0.000011390625 $$

Step4: Compute $(1-p)^{n-k}$ term

$$ (1-p)^{n-k} = 0.85^4 = 0.52200625 $$

Step5: Multiply all terms

$$ P(6) = 210 \times 0.000011390625 \times 0.52200625 $$

$$ P(6) = 210 \times 0.0000059459033203125 = 0.001248639697265625 $$

Step6: Round to nearest thousandth

$$ P(6) \approx 0.001 $$

Answer:

0.001