Question

1. a student takes a 20 question test and has an 80% chance of getting any question correct. let x = the number of questions the students answers correctly. find:

a. p(x = 18)
b. p(x ≤ 15)
c. p(x ≥ 17)
d. p(x = 20)
e. p(x < 18)
f. p(x > 10)

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 \(n\) 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, \(n = 20\), \(p=0.8\), \(1 - p = 0.2\)

Step2: Calculate \(P(X = 18)\)

\(C(20,18)=\frac{20!}{18!(20 - 18)!}=\frac{20\times19}{2\times1}=190\)
\(P(X = 18)=C(20,18)\times(0.8)^{18}\times(0.2)^{2}\)
\(P(X = 18)=190\times(0.8)^{18}\times(0.2)^{2}\approx0.1369\)

Step3: Calculate \(P(X\leq15)\)

\(P(X\leq15)=1-(P(X = 16)+P(X = 17)+P(X = 18)+P(X = 19)+P(X = 20))\)
\(C(20,16)=\frac{20!}{16!(20 - 16)!}=\frac{20\times19\times18\times17}{4\times3\times2\times1}=4845\)
\(P(X = 16)=C(20,16)\times(0.8)^{16}\times(0.2)^{4}=4845\times(0.8)^{16}\times(0.2)^{4}\approx0.2182\)
\(C(20,17)=\frac{20!}{17!(20 - 17)!}=\frac{20\times19\times18}{3\times2\times1}=1140\)
\(P(X = 17)=C(20,17)\times(0.8)^{17}\times(0.2)^{3}=1140\times(0.8)^{17}\times(0.2)^{3}\approx0.2702\)
\(P(X = 18)\approx0.1369\) (calculated above)
\(C(20,19)=\frac{20!}{19!(20 - 19)!}=20\)
\(P(X = 19)=C(20,19)\times(0.8)^{19}\times(0.2)^{1}=20\times(0.8)^{19}\times(0.2)^{1}\approx0.2702\)
\(C(20,20)=\frac{20!}{20!(20 - 20)!}=1\)
\(P(X = 20)=(0.8)^{20}\approx0.0115\)
\(P(X\leq15)=1-(0.2182 + 0.2702+0.1369+0.2702 + 0.0115)\approx0.093\)

Step4: Calculate \(P(X\geq17)\)

\(P(X\geq17)=P(X = 17)+P(X = 18)+P(X = 19)+P(X = 20)\)
\(P(X\geq17)=0.2702+0.1369+0.2702 + 0.0115\approx0.6888\)

Step5: Calculate \(P(X = 20)\)

\(P(X = 20)=(0.8)^{20}\approx0.0115\)

Step6: Calculate \(P(X\lt18)\)

\(P(X\lt18)=1 - P(X\geq18)=1-(P(X = 18)+P(X = 19)+P(X = 20))\)
\(P(X\lt18)=1-(0.1369+0.2702 + 0.0115)\approx0.5814\)

Step7: Calculate \(P(X\gt10)\)

\(P(X\gt10)=1-(P(X = 0)+P(X = 1)+\cdots+P(X = 10))\)
Using a binomial calculator or software (since calculating each term \(P(X = k)\) for \(k = 0\) to \(10\) is time - consuming), \(P(X\gt10)\approx0.9999\)

Answer:

A. \(P(X = 18)\approx0.1369\)
B. \(P(X\leq15)\approx0.093\)
C. \(P(X\geq17)\approx0.6888\)
D. \(P(X = 20)\approx0.0115\)
E. \(P(X\lt18)\approx0.5814\)
F. \(P(X\gt10)\approx0.9999\)