The Professor's Response

Hello Professor,

I need help with this question, "What is the probability that when making random guesses on a fifty question multiple choice exam, with four answer choices for each question, that you miss only thirty-five of the questions?"

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Professor McGukian

Hi Katherine,

The problem is asking, "what is the probability...," so, in order to answer this question, we first need to figure out what method of probability is required. There is more than one selection involved because we must select 50 answers to complete the exam. This indicates the problem is likely multiplication rule or it is binomial probability. It looks like binomial probability because we want the probability that only 35 of the guesses are incorrect. The basic multiplication rule would have all of the guesses turn out the same way, so either all correct or all incorrect. We can check the criteria for a binomial problem to be sure.

Does the problem have a fixed number of trials?

Yes, there are 50 guesses. This means our n = 50.

Does the outcome for each guess fall into one of two categories?

Yes, there are only two possible outcomes, correct or incorrect.

Are the guesses independent?

Yes, one guess does not affect the other.

Is there a constant probability of guessing incorrectly?

Yes, since each question has just four answer choices, we can say that for each guess there is a 3/4 chance of guessing incorrectly. This means that our p = 3/4 

Ok, since this is a binomial experiment, let's define x as the number of incorrect guesses.

x = # of incorrect guesses

Then we want the probability that x is equal to 35, P(x = 35).

Recall that our n = 50 and our p = 3/4 = 0.75. Then we can find q = 1 - p = 1 - 3/4 = 1/4 = 0.25.

Finally, we can use our binomial formula: P(x = 35) = 50C35* (.75)^35*(.25)^15 

Notice that our x is defined as the number of wrong answers, so our p must be the probability of a wrong answer, and it must be raised to the 35th power since that is how many wrong guesses we want.

I hope that helps,

Professor McGuckian   

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