# Succeed with one multi-choice test out of five if answers are guessed

I'm taking a class in probability, and I have some issues understanding some concepts. Thing is, I am trying to calculate the probability that you can succeed with at least one multi-choice test out of five, if all the answers are guessed. In each test there are 18 questions, and 9 must be answered correctly for the test to be passed. Each question has 4 choices, so the chance of getting a question right is $1 \over 4$.

I figured that since 9 of the 18 questions had to be correctly answered for the test to succeed, the probability of succeeding with one test, P(success), would be

$P(success) = {1 \over {4^9}}$

for each test. The probability of failing one test, P(success*), would then be

$P(success*) = 1 - P(success) = 1 - {1 \over {4^9}}$

Thus, the probability to fail every test would be

$P(success*)^5 = (1- {1 \over {4^9}})^5$

Then, finally, the probability to succeed with at least one of the five tests, P(at least one success), would be

$P(at\ least\ one\ success) = 1-P(success*)^5 = 1-(1-{1 \over {4^9}})^5 \approx 0.000019$

This, however, is not even close to being correct.

I also tried using the number of combinations of 9 successes and 9 fails for P(success):

$P(success) = {18 \choose 9}({1 \over4})^9(1-{1 \over 4})^{18-9}$

Which then gives

$P(success*) = 1-P(success) = 1-({18 \choose 9}({1 \over4})^9(1-{1 \over 4})^{18-9})$

and thusly

$P(at\ least\ one\ success) = 1 - P(success*)^5 = 1 - (1-({18 \choose 9}({1 \over4})^9(1-{1 \over 4})^{18-9}))^5 \approx 0.0677$

This is closer, but still not correct. The answer provided by the book is just

$P(succeed\ with\ one\ test) = 1-P(fail\ every\ test) = 1 - 0.98065^5 \approx 0.094$

Since I know that $1- P(success*)^5$ is correct, I understand that is is $P(success*)$ that is wrong, but I don't know how and why it's wrong. So I would really appreciate any help with understanding this.

-

Your first go at it, $P(success)=(1/4)^9$, is not correct. This would give the probability that you answered, e.g., the first $9$ questions of a particular test correctly. It does not take into account how you fared on the other $9$ questions. If you wanted the probability that you answered the first $9$ correctly and the others incorrectly, it would be given by $(1/4)^9(3/4)^9$.

But there are many ways in which you could answer exactly $9$ questions correctly. In fact, there are $18 \choose 9$ ways in which this can occur. So, your second go at it, $P(success) = {18 \choose 9}({1 \over4})^9(1-{1 \over 4})^{18-9}$, is closer. This correctly gives the probability that you answered exactly $9$ questions correctly.

But, this is still not correct. You pass the test if you answered at least $9$ questions correctly. So you need to find the probability that you answered exactly $9$, or exactly $10$, or ...., or exactly $18$ correctly. Then add those up.

The probability that you failed a particular test is the probability that you answered at most $8$ questions correctly. Calculating this would be a bit more direct: $$P(\text{ failed a particular test})=\sum_{i=0}^8 P(\text{answered exactly }i\text{ questions correctly})=\sum_{i=1}^8 {18\choose i} (1/4)^i (3/4)^{18-i}.$$ The above is just the cumulative distribution function of a Binomial variable with $18$ trials and success factor $1/4$ evaluated at $8$. There are online calculators, such as this, that will compute it. From the link, we have $P(\text{failed a particular test})\approx .98065222$.

So, the probability that you passed at least one test is $1-P(\text{ failed all five})\approx 1-(.98065)^5$.

-
Thank you very much, I can see how this makes sense! – Psyberion Jul 7 '13 at 15:42
@Psyberion You're welcome; glad to help. – David Mitra Jul 7 '13 at 15:49