There are two questions with 4 options each. What are the chances of getting both answers correct?
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Answer:
To compute a probability, we need to decide on a random process and a model to describe that process. Your question doesn't tell me anything about a random process at all so you haven't specified enough to really nail down what model to use. I'll make my own assumptions that I hope agree with what you intended to ask.
First, I'll assume that each of the five questions has four choices (call them a, b, c, and d), and that exactly one of these four is correct. Next, I'll assume that on each question, a person picks one of the four choices at random in such a way that:
Each of the four choices is equally likely to be picked on each question.
The choices made on each question are independent. (This roughly means that knowing what got picked on some group of questions doesn't impact what gets chosen on some other questions.)
The second bulleted assumption allows us to ignore how the quiz author decided which letter to assign to the correct answer. They could all just be assigned to a (for example) and it wouldn't impact this calculation. Equivalently, we could instead assume that the author chose which letter to assign to the correct answer independently on each problem with an equal probability for each choice. This assumption would yield the same answer regardless of how the student decided which letter to guess.
With these assumptions in hand, it turns out that this random process is a Binomial Experiment. The number of correctly guessed answers (call it XX) follows a Binomial distribution with parameters n=5n=5 and p=14p=14. For any k∈{0,1,2,3,4,5}k∈{0,1,2,3,4,5} it turns out that the probability that X=kX=k is given by:
P(X=k)=(nk)pk(1−p)n−kP(X=k)=(nk)pk(1−p)n−k
Plugging in k=4k=4 correctly guessed answers, n=5n=5 total questions, and p=14p=14 chance to guess each right yields:
P(X=4)=(54)(14)4(34)5−4=151024<1.5P(X=4)=(54)(14)4(34)5−4=151024<1.5%
It should be noted that the assumptions are really important. In my life, I've taken many quizzes like the one you describe, and I got 4 (or more) right on MOST of them so the probability of 1.5% certainly isn't "true" for my approach to test taking.
But even if you do use a random guess strategy and a random assignment of letter to the correct answer, the independence is essential.
For example, the quiz author could choose at random with equal probability which letter to assign to the correct answer on the first question. She could then just follow a sequential pattern for the rest of the answers. (For example, if she randomly chose b for question 1, then she would choose c for question 2, d for question 3, a for questions 4, and back to b for question 5.) It turns out that with such an assignment, on each question, there is a 1/4 chance of each letter being the correct answer; however, the answers are not independent. Similarly, the student could randomly pick a guess for the first question (equally likely to pick any answer) but then follow a sequential pattern for the remaining guesses (just like the author of the quiz).
In this scenario, we still have a random assignment of correct answers with all choices equally likely, and we still have a random guess with all choices equally likely to be guessed. However, there are now only two possible scenarios. Either the guess and answer match on the first question (and therefore on EVERY subsequent question) or they don't match on the first question (and therefore don't math on ANY subsequent answer). In this scenario, there is a 25% chance that the student guesses all 5 answers right and a 75% chance that the student guesses no answer correctly.
As you can see, the assumption of independence makes a big difference.
Step-by-step explanation: