Question

assume that a procedure yields a binomial distribution with ( n = 2 ) trials and a probability of success of ( p = 0.40 ). find the probability that the number of successes ( x ) is exactly 1.
( p(1)=square ) (round to three decimal places as needed.)

Explanation:

Step1: Recall binomial probability formula

The binomial probability formula is \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(C(n,k)=\frac{n!}{k!(n - k)!}\), \(n\) is the number of trials, \(k\) is the number of successes, and \(p\) is the probability of success in a single - trial. Here, \(n = 2\), \(k = 1\), \(p=0.4\), and \(1-p = 0.6\).

Step2: Calculate the combination \(C(n,k)\)

First, calculate \(C(2,1)=\frac{2!}{1!(2 - 1)!}=\frac{2!}{1!1!}=\frac{2\times1!}{1!×1!}=2\).

Step3: Substitute values into the binomial formula

Substitute \(n = 2\), \(k = 1\), \(p = 0.4\), and \(1 - p=0.6\) into the formula \(P(X = 1)=C(2,1)\times(0.4)^{1}\times(0.6)^{2 - 1}\).
\(P(X = 1)=2\times0.4\times0.6\).

Step4: Calculate the result

\(P(X = 1)=2\times0.24 = 0.48\).

Answer:

\(0.480\)