Question

find the probability of exactly three successes in eight trials of a binomial experiment in which the probability of success is 45%.
n = 8 p = 0.45 q = 0.55 k = 3
enter values into the formula.
p(k) = _nc_k(p)^k(q)^{n - k}
p(?) = _c_( )^( )^{ - }

Explanation:

Step1: Substitute values into the formula

We are given \(n = 8\), \(p=0.45\), \(q = 0.55\), \(k = 3\).
The binomial probability formula is \(P(k)={}_{n}C_{k}(p)^{k}(q)^{n - k}\).
Substituting the values, we get \(P(3)={}_{8}C_{3}(0.45)^{3}(0.55)^{8 - 3}\).

Step2: Calculate the combination

The combination formula is \({}_{n}C_{k}=\frac{n!}{k!(n - k)!}\).
For \(n = 8\) and \(k = 3\), \({}_{8}C_{3}=\frac{8!}{3!(8 - 3)!}=\frac{8!}{3!5!}=\frac{8\times7\times6\times5!}{3\times2\times1\times5!}=56\).

Step3: Calculate the powers

\((0.45)^{3}=0.45\times0.45\times0.45 = 0.091125\)
\((0.55)^{5}=0.55\times0.55\times0.55\times0.55\times0.55\approx0.0503284375\)

Step4: Calculate the probability

\(P(3)=56\times0.091125\times0.0503284375\)
\(P(3)\approx56\times0.00458128\)
\(P(3)\approx0.256552\)

Answer:

\(P(3)={}_{8}C_{3}(0.45)^{3}(0.55)^{8 - 3}\approx0.2566\) (rounded to four decimal places)