Question

lucia created a marshmallow launching machine. she is confident that 77% of the time, a launched marshmallow will travel at least 2 meters. assuming lucia is correct, if she launches 4 marshmallows, what is the probability that fewer than 3 will travel at least 2 meters? write your answer as a decimal rounded to the nearest thousandth.

Explanation:

Step1: Identify the binomial probability formula

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

Here, \(n = 4\), \(p=0.77\), and we want to find \(P(X\lt3)=P(X = 0)+P(X = 1)+P(X = 2)\)

Step2: Calculate \(P(X = 0)\)

\(C(4,0)=\frac{4!}{0!(4 - 0)!}=1\)
\(P(X = 0)=C(4,0)\times(0.77)^{0}\times(1 - 0.77)^{4-0}\)
\(=1\times1\times(0.23)^{4}\)
\(=1\times1\times0.00279841=0.00279841\)

Step3: Calculate \(P(X = 1)\)

\(C(4,1)=\frac{4!}{1!(4 - 1)!}=\frac{4!}{1!3!}=4\)
\(P(X = 1)=C(4,1)\times(0.77)^{1}\times(1 - 0.77)^{4 - 1}\)
\(=4\times0.77\times(0.23)^{3}\)
\(=4\times0.77\times0.012167=4\times0.00936859 = 0.03747436\)

Step4: Calculate \(P(X = 2)\)

\(C(4,2)=\frac{4!}{2!(4 - 2)!}=\frac{4\times3\times2!}{2!\times2!}=6\)
\(P(X = 2)=C(4,2)\times(0.77)^{2}\times(1 - 0.77)^{4-2}\)
\(=6\times(0.77)^{2}\times(0.23)^{2}\)
\(=6\times0.5929\times0.0529\)
\(=6\times0.03136441=0.18818646\)

Step5: Calculate \(P(X\lt3)\)

\(P(X\lt3)=P(X = 0)+P(X = 1)+P(X = 2)\)
\(P(X\lt3)=0.00279841+0.03747436 + 0.18818646\)
\(P(X\lt3)=0.22845923\approx0.228\)

Answer:

\(0.228\)