Question

Please help!!

A multiple-choice test consists of 28 questions with possible answers of a,b,c,d. Estimate the probability that with random guessing, the number of correct answers is at least 12.

Answer 1

Approximately 0.0294.

Explanation:

Assume that there's only one correct choice in each question.

• The chance of getting a question correct by random guess is 1/4.
• The chance of getting a question wrong by random guess is 3/4.

What's the probability that exactly 12 answers are correct?

• 12 out of the 28 answers need to be correct.
  `\displaystyle \left((1)/(4)\right)^(12)`.
• The other 28 - 12 answers need to be incorrect. Multiply by
  `\displaystyle \left((3)/(4)\right)^(28 - 12)`.
• There are more than one way of choosing 12 answers out of 28 without an order. Multiply by the combination "12-choose-28"
  `\displaystyle \left(\begin{array}{c}12\\28\end{array}\right)`.

The probability of getting exactly 12 answers correct is:

`\displaystyle \left((1)/(4)\right)^(12) * \left((3)/(4)\right)^(28 - 12)* \left(\begin{array}{c}28\\12\end{array}\right)\approx 0.0182`.

With the same logic, the probability of getting
`x` (
`x\in \mathbb{Z}`,
`12\le x\le 28`) correct out of the 28 random answers will be

`\displaystyle \left((1)/(4)\right)^(x) * \left((3)/(4)\right)^(28 - x)* \left(\begin{array}{c}28\\x\end{array}\right)`.

The probability of getting at least 12 correct out of 28 random answers is the sum of

• the probability of getting exactly 12 correct out of 28, plus
• the probability of getting exactly 13 correct out of 28, plus
• the probability of getting exactly 14 correct out of 28, plus
• the probability of getting exactly 15 correct out of 28, plus
• the probability of getting exactly 16 correct out of 28, plus
• ... all the way to the probability of getting exactly 28 correct out of 28.

The Sigma notation might help:

`\displaystyle \sum_(x = 12)^(28){\left[\left((1)/(4)\right)^(x) * \left((3)/(4)\right)^(28 - x)* \left(\begin{array}{c}x\\28\end{array}\right)\right]}`.

Evaluate this sum (preferably with a calculator)

`\displaystyle \sum_(x = 12)^(28){\left[\left((1)/(4)\right)^(x) * \left((3)/(4)\right)^(28 - x)* \left(\begin{array}{c}x\\28\end{array}\right)\right]} \approx 0.0294`.