Question

give your answers rounded upto 3 decimal places. examples;
0.2478749077 should be 0.248
0.2471749077 should be 0.247
if x is binomial with n = 20 and p =.6, find the probability that
x is at least 10
x equals 12
x is at most 12
x is greater than or equal to 9 but not more than 11

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 \(C(n,k)=\frac{n!}{k!(n - k)!}\)

Step2: Calculate \(P(X\geq10)\)

\(P(X\geq10)=\sum_{k = 10}^{20}C(20,k)\times0.6^{k}\times0.4^{20 - k}\)
Using a binomial probability calculator or software (e.g., in R: pbinom(9,20,0.6,lower.tail = FALSE)), we get \(P(X\geq10)\approx0.584\)

Step3: Calculate \(P(X = 12)\)

\(C(20,12)=\frac{20!}{12!(20 - 12)!}=\frac{20\times19\times\cdots\times9}{8\times7\times\cdots\times1}=125970\)
\(P(X = 12)=C(20,12)\times0.6^{12}\times0.4^{8}\)
\(P(X = 12)=125970\times(0.6)^{12}\times(0.4)^{8}\approx0.180\)

Step4: Calculate \(P(X\leq12)\)

\(P(X\leq12)=\sum_{k = 0}^{12}C(20,k)\times0.6^{k}\times0.4^{20 - k}\)
Using a binomial probability calculator or software (e.g., in R: pbinom(12,20,0.6)), we get \(P(X\leq12)\approx0.584\)

Step5: Calculate \(P(9\leq X\leq11)\)

\(P(9\leq X\leq11)=P(X = 9)+P(X = 10)+P(X = 11)\)
\(C(20,9)=\frac{20!}{9!(20 - 9)!}=167960\), \(C(20,10)=\frac{20!}{10!(20 - 10)!}=184756\), \(C(20,11)=\frac{20!}{11!(20 - 11)!}=167960\)
\(P(X = 9)=C(20,9)\times0.6^{9}\times0.4^{11}\)
\(P(X = 10)=C(20,10)\times0.6^{10}\times0.4^{10}\)
\(P(X = 11)=C(20,11)\times0.6^{11}\times0.4^{9}\)
\(P(9\leq X\leq11)\approx0.251\)

Answer:

• \(P(X\geq10)\approx0.584\)
• \(P(X = 12)\approx0.180\)
• \(P(X\leq12)\approx0.584\)
• \(P(9\leq X\leq11)\approx0.251\)