Question

suppose you are taking an exam with 12 questions and you are required to get 8 or more right answers to pass. complete parts a through d below.
a. with a 12 - question true/false test, what is the probability of getting at least 8 answers correct by guessing?
the probability of getting at least 8 answers correct is
(type an integer or decimal rounded to three decimal places as needed.)
b. with a 12 - question multiple - choice test where there are four possible choices for each question, what is the probability of getting at least 8 answers correct by guessing? only one of the choices is correct for each question.
the probability of getting at least 8 answers correct by guessing is
(type an integer or decimal rounded to three decimal places as needed.)
c. with a 12 - question multiple - choice test where there are five possible choices for each question, what is the probability of getting at least 8 answers correct by guessing? only one of the choices is correct for each question.
the probability of getting at least 8 answers correct by guessing is
(type an integer or decimal rounded to three decimal places as needed.)
d. which test (of those described in parts a, b, and c in this exercise) would be easiest to pass by guessing, which would be hardest, and why?
the
would be easiest to pass by guessing, since the probability of getting at least 8 answers correct by guessing on that test is
the
would be hardest to pass by guessing, since the probability of getting at least 8 answers correct by guessing on that test is

Explanation:

Step1: 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 \(C(n,k)=\frac{n!}{k!(n - k)!}\). For getting at least \(8\) correct answers, we need to calculate \(P(X\geq8)=P(X = 8)+P(X = 9)+P(X = 10)+P(X = 11)+P(X = 12)\)

Part a: True/False test (\(p=\frac{1}{2}\), \(n = 12\))

- **For \(k = 8\)**:
\(C(12,8)=\frac{12!}{8!(12 - 8)!}=\frac{12\times11\times10\times9}{4\times3\times2\times1}=495\)
\(P(X = 8)=C(12,8)\times(\frac{1}{2})^{8}\times(1-\frac{1}{2})^{12 - 8}=495\times(\frac{1}{2})^{12}\)
- **For \(k = 9\)**:
\(C(12,9)=\frac{12!}{9!(12 - 9)!}=\frac{12\times11\times10}{3\times2\times1}=220\)
\(P(X = 9)=C(12,9)\times(\frac{1}{2})^{9}\times(1-\frac{1}{2})^{12 - 9}=220\times(\frac{1}{2})^{12}\)
- **For \(k = 10\)**:
\(C(12,10)=\frac{12!}{10!(12 - 10)!}=\frac{12\times11}{2\times1}=66\)
\(P(X = 10)=C(12,10)\times(\frac{1}{2})^{10}\times(1-\frac{1}{2})^{12 - 10}=66\times(\frac{1}{2})^{12}\)
- **For \(k = 11\)**:
\(C(12,11)=\frac{12!}{11!(12 - 11)!}=12\)
\(P(X = 11)=C(12,11)\times(\frac{1}{2})^{11}\times(1-\frac{1}{2})^{12 - 11}=12\times(\frac{1}{2})^{12}\)
- **For \(k = 12\)**:
\(C(12,12)=\frac{12!}{12!(12 - 12)!}=1\)
\(P(X = 12)=C(12,12)\times(\frac{1}{2})^{12}\times(1-\frac{1}{2})^{12 - 12}=1\times(\frac{1}{2})^{12}\)

\(P(X\geq8)=(495 + 220+66 + 12+1)\times(\frac{1}{2})^{12}=\frac{794}{4096}\approx0.194\)

Part b: Multiple - choice test with \(4\) options (\(p=\frac{1}{4}\), \(n = 12\))

- **For \(k = 8\)**:
\(C(12,8)=\frac{12!}{8!(12 - 8)!}=495\)
\(P(X = 8)=C(12,8)\times(\frac{1}{4})^{8}\times(1-\frac{1}{4})^{12 - 8}=495\times(\frac{1}{4})^{8}\times(\frac{3}{4})^{4}\)
\(P(X = 8)=495\times\frac{1}{65536}\times\frac{81}{256}\approx495\times\frac{81}{16777216}\approx0.0024\)
- **For \(k = 9\)**:
\(C(12,9)=\frac{12!}{9!(12 - 9)!}=220\)
\(P(X = 9)=C(12,9)\times(\frac{1}{4})^{9}\times(\frac{3}{4})^{3}=220\times\frac{1}{262144}\times\frac{27}{64}\approx220\times\frac{27}{16777216}\approx0.0004\)
- **For \(k = 10\)**:
\(C(12,10)=\frac{12!}{10!(12 - 10)!}=66\)
\(P(X = 10)=C(12,10)\times(\frac{1}{4})^{10}\times(\frac{3}{4})^{2}=66\times\frac{1}{1048576}\times\frac{9}{16}\approx66\times\frac{9}{16777216}\approx0.000035\)
- **For \(k = 11\)**:
\(C(12,11)=\frac{12!}{11!(12 - 11)!}=12\)
\(P(X = 11)=C(12,11)\times(\frac{1}{4})^{11}\times(\frac{3}{4})^{1}=12\times\frac{1}{4194304}\times\frac{3}{4}\approx12\times\frac{3}{16777216}\approx0.0000021\)
- **For \(k = 12\)**:
\(C(12,12)=\frac{12!}{12!(12 - 12)!}=1\)
\(P(X = 12)=C(12,12)\times(\frac{1}{4})^{12}\times(\frac{3}{4})^{0}=\frac{1}{16777216}\approx0.00000006\)

\(P(X\geq8)\approx0.0024 + 0.0004+0.000035 + 0.0000021+0.00000006\approx0.003\)

Part c: Multiple - choice test with \(5\) options (\(p=\frac{1}{5}\), \(n = 12\))

- **For \(k = 8\)**:
\(C(12,8)=\frac{12!}{8!(12 - 8)!}=495\)
\(P(X = 8)=C(12,8)\times(\frac{1}{5})^{8}\times(1-\frac{1}{5})^{12 - 8}=495\times(\frac{1}{5})^{8}\times(\frac{4}{5})^{4}\)
\(P(X = 8)=495\times\frac{1}{390625}\times\frac{256}{625}\approx495\times\frac{256}{244140625}\approx0.00052\)
- **For \(k = 9\)**:
\(C(12,9)=\frac{12!}{9!(12 - 9)!}=220\)
\(P(X = 9)=C(12,9)\times(\frac{1}{5})^{9}\times(\frac{4}{5})^{3}=220\times\frac{1}{1953125}\times\frac{64}{125}\approx220\times\frac{64}{244140625}\approx0.000058\)
- **For \(k = 10\)**:
\(C(12,10)=\frac{12!}{10!(12 - 10)!}=66\)
\(P(X = 10)=C(12,10)\times(\frac{1}{5})^{10}\times(\frac{4}{5})^{2}=66\times\frac{1}{9765625}\times\frac{16}{25}\approx66\times\frac{16}{244140625}\approx0.0000043\)
- **For \(k = 11\)**:
\(C(12,11)=\frac{12!}{11!(12 - 11)!}=12\)
\(P(X = 11)=C(12,11)\times(\frac{1}{5})^{11}\times(\frac{4}{5})^{1}=12\times\frac{1}{48828125}\times\frac{4}{5}\approx12\times\frac{4}{244140625}\approx0.0000002\)
- **For \(k = 12\)**:
\(C(12,12)=\frac{12!}{12!(12 - 12)!}=1\)
\(P(X = 12)=C(12,12)\times(\frac{1}{5})^{12}\times(\frac{4}{5})^{0}=\frac{1}{244140625}\approx0.000000004\)

\(P(X\geq8)\approx0.00052+0.000058 + 0.0000043+0.0000002+0.000000004\approx0.001\)

Answer:

a. \(0.194\)
b. \(0.003\)
c. \(0.001\)
d. The true/false test would be easiest to pass by guessing, since the probability of getting at least \(8\) answers correct by guessing on that test is the highest. The multiple - choice test with five options would be hardest to pass by guessing, since the probability of getting at least \(8\) answers correct by guessing on that test is the lowest.