The Notes: A quiz consists of 10 multiple-choice questions, each with 4 possible answers. For someone who makes random guesses for all of the answers, find...

Wednesday, November 13, 2013

A quiz consists of 10 multiple-choice questions, each with 4 possible answers. For someone who makes random guesses for all of the answers, find...

This is a great question, and one that you will probably see several times in a Probability and Statistics class.

The main thing you will need is the Binomial Probability formula, shown below:

$\displaystyle{P}{\left({k}\right)}={n}{C}{k}\cdot{{p}}^{{k}}\cdot{{\left({1}-{p}\right)}}^{{{n}-{k}}}$

The first part is “n choose k”, where n is the total number of questions and k is the number of questions you want to consider. So you are starting by finding all the possible combinations of choosing 7 random questions out of the 10. That is then multiplied by the probability of getting a question right 7 times and by the probability of getting a question wrong 3 times.

So, to start, let’s consider the probability you need to find. You want to find the probability of all possibilities of passing the quiz, which includes scores of 70%, 80%, 90%, and 100% totaled. This is where you will need the Binomial Probability formula.

For our problem, n is the number of questions, 10, and k is the number of correct answers, 7 through 10. The probability p of getting a question right is and the probability of getting a question wrong is or .

So, the probability of getting a score of exactly 70% is $\displaystyle{10}{C}{7}\cdot{{\left(\frac{{1}}{{4}}\right)}}^{{7}}\cdot{{\left(\frac{{3}}{{4}}\right)}}^{{{3}}}$

The probability of getting a score of exactly 80% is $\displaystyle{10}{C}{8}\cdot{{\left(\frac{{1}}{{4}}\right)}}^{{8}}\cdot{{\left(\frac{{3}}{{4}}\right)}}^{{{2}}}$

The probability of getting a score of exactly 90% is $\displaystyle{10}{C}{9}\cdot{{\left(\frac{{1}}{{4}}\right)}}^{{9}}\cdot{{\left(\frac{{3}}{{4}}\right)}}^{{{1}}}$

The probability of getting a score of exactly 100% is $\displaystyle{10}{C}{10}\cdot{{\left(\frac{{1}}{{4}}\right)}}^{{10}}\cdot{{\left(\frac{{3}}{{4}}\right)}}^{{{0}}}$

We now just add up all these probabilities for our answer:

$\displaystyle{P}{\left(\ge{70}%\right)}={10}{C}{7}\cdot{{\left(\frac{{1}}{{4}}\right)}}^{{7}}\cdot{{\left(\frac{{3}}{{4}}\right)}}^{{{3}}}+{10}{C}{8}\cdot{{\left(\frac{{1}}{{4}}\right)}}^{{8}}\cdot{{\left(\frac{{3}}{{4}}\right)}}^{{{2}}}+{10}{C}{9}\cdot{{\left(\frac{{1}}{{4}}\right)}}^{{9}}\cdot{{\left(\frac{{3}}{{4}}\right)}}^{{{1}}}+{10}{C}{10}\cdot{{\left(\frac{{1}}{{4}}\right)}}^{{10}}\cdot{{\left(\frac{{3}}{{4}}\right)}}^{{{0}}}$