Question

Suppose you take a 12-question true or false quiz by guessing each answer. Use the binomial table to find the probability of guessing 6 or more questions correctly.

Answer 1

0.6127 = 61.27% probability of guessing 6 or more questions correctly.

Explanation:

For each question, there are only two possible outcomes. Either you guess the correct answer, or you do not. The probability of guessing the correct answer of a question is independent of other questions. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

In which
`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

12 questions:

This means that
`n = 12`

True-false:

Two options, one of which is correct. So
`p = \frac{1}[2} = 0.5`

Find the probability of guessing 6 or more questions correctly.

`P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)`

In which

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 6) = C_(12,6).(0.5)^(6).(0.5)^(6) = 0.2256`

`P(X = 7) = C_(12,7).(0.5)^(7).(0.5)^(5) = 0.1934`

`P(X = 8) = C_(12,8).(0.5)^(8).(0.5)^(4) = 0.1208`

`P(X = 9) = C_(12,9).(0.5)^(9).(0.5)^(3) = 0.0537`

`P(X = 10) = C_(12,10).(0.5)^(10).(0.5)^(2) = 0.0161`

`P(X = 11) = C_(12,11).(0.5)^(11).(0.5)^(1) = 0.0029`

`P(X = 12) = C_(12,12).(0.5)^(12).(0.5)^(0) = 0.0002`

`P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.2256 + 0.1934 + 0.1208 + 0.0537 + 0.0161 + 0.0029 + 0.0002 = 0.6127`

0.6127 = 61.27% probability of guessing 6 or more questions correctly.