Question

check here for instructional material to complete this problem. evaluate $_nc_xp^x(1 - p)^{n - x}$ for n = 6, p = 0.1, x = 2. the answer is . (round to four decimal places as needed.)

Explanation:

Step1: Calculate combination formula

$_{n}C_{x}=\frac{n!}{x!(n - x)!}$, where $n = 6$ and $x=2$. So $_{6}C_{2}=\frac{6!}{2!(6 - 2)!}=\frac{6!}{2!4!}=\frac{6\times5}{2\times1}=15$.

Step2: Calculate $(1 - p)^{n - x}$

Given $p = 0.1$, then $1-p=0.9$, and $n - x=6 - 2 = 4$. So $(1 - p)^{n - x}=0.9^{4}=0.6561$.

Step3: Calculate $p^{x}$

Given $p = 0.1$ and $x = 2$, so $p^{x}=0.1^{2}=0.01$.

Step4: Calculate the final result

Multiply the three - results together: $_{n}C_{x}p^{x}(1 - p)^{n - x}=15\times0.01\times0.6561 = 0.098415\approx0.0984$.

Answer:

$0.0984$