Thread: Probability, should be really easy..

Hello someone please help me asap, and show me exactly what todo!!

A class of 12 students take a multiple choice test consisting of 8 questions with answers from A to D. If every student guesses every answer, find the probability that less than 4 students pass. (A pass is 4 or more questions correct.)

I thought that the questions ranging from A – D has nothing todo with it, there is a population of 12, 8 questions so number of trials is 8, then the probability associated to this is 50% because you need 4 or more correct to pass, as 4 is half of 8 hence 50%. Firstly is hits correct, secondly what mathematical formulas will I need to solve this, and how?? Is it just a simple binomial, or is it a normal distributed thing, I don’t know

Thank you, much appreciated!!

The fact that the questions range from A-D is important. Before you calculate the probability of less than 4 students passing you have to figure out the probability of 1 student passing. This can be worked out as follows:

Because there are 4 choices on the test A-D, a student has a 1/4 (or 0.25) chance of getting a question correct if they guess. A student passes if he gets 4 or more of the 8 questions correctly. So this is a binomial with no of questions n=8, and p=0.25. Let X be the number of questions a student answers correctly. The probability that the student passes is equal to the Pr(student passes) = Pr(X=4) + Pr(X=5) + Pr(X=6) + Pr(X=7) + Pr(X=8) or alternately you can calculate 1 - Pr(student fails) which means the student gets less than 4 on the test. This will save a bit of time because you only have to calculate 1 - Pr(X=3) - Pr(X=2) - Pr(X=1) - Pr(X=0).

Now Pr(X=x) = (n C x) * (p^x) * (1-p)^(n-x)
So Pr(X=3) = (8!)/(3!*5!) * (0.25)^3 * (0.75)^5 = 0.20764 (I recommend working it out as a fraction or keeping a few decimal places so you don't lose too much when rounding the number)

So if you do the other calculations for Pr(X=2), Pr(X=1) and Pr(X=0) and subtract them all from 1 you will work out the probabiltity that 1 student will pass. Then you have a new binomial distribution with number of students = 12 and probability of passing = p (which you just calculated). So if you let Y = the number of students who pass then the probability that less than 4 students pass is equal to Pr(Y=3) + Pr(Y=2) + Pr(Y=1) + Pr(Y=0).

Hope this helps. It's a slightly tricky question

So would the answer be, there is an 11% chance that a student passes and 96% chance that less than 4 passes?

Yep, nice work