# Probability to answer all question of an exam correctly given 5 choices

Suppose a teacher gives his class a set of $$10$$ questions with the information that the final exam will consist of a random selection of $$5$$ of them. If a student has figured out how to solve $$7$$ of them, what is the probability that he answers all of the questions correctly? how about at least $$4$$ of the problems?

### Attempt:

First, $${10 \choose 5}$$ is the size of the sample space. Since it is given that he can do $$7$$ problems, then probaility that he answer a problem correctly must be $$\frac{7}{10} = 0.7$$.

Thus, probability that he answers all questions correctly is

$$P(Q1 \; correct \; \; AND \; \; Q2 \; correct \; \; AND ... AND \; \; Q5 \; correct ) = 0.7 \times 0.7 \times ... \times 0.7 = 0.7^5$$

Now, the probability that he answer a question wrong is $$1 - 0.7 = 0.3$$.

Thus,

$$P(\text{at least 4 correct}) = P(\text{at most one wrong}) = P(\text{no wrong}) + P(\text{1 wrong}) = 0.7^5+0.3$$

But, I feel we we dont really need to know the size of sample space. Or, perhaps Im missing something here?

• You determined the number of possible combinations correctly. To find the number of successful combinations, consider that all $5$ questions must be among the $7$ the student can solve. Apr 17 '18 at 16:41
• Given that he knows how to do 7 of the problems, the number of ways that 5 of those 7 problems can appear on the test is $7 \choose 5$. Your sample size is correct. Apr 17 '18 at 16:42

There are $_7C_5 = 21$ ways to choose $5$ questions among the $7$ he knows how to solve. There are $_{10}C_5 = 252$ ways to choose $5$ questions from the $10$.

So, the probability of getting all $5$ right is $P_5 = 21/252 \doteq 0.083$.

For the second part, we need to add in the number of ways to get exactly $4$ questions that he knows how to do.

So, choose $4$ he knows how to do ($_7C_4 = 35$) and one he doesn't ($3$) for a total of $35 \cdot 3 = 105$ combinations.

Now, the probability of getting at least $4$ right is $$P_5 + P_4 = 21/252 + 105/252 = 126/252 = 0.5.$$

• it is unclear how you get the second probaility. why you adding? Apr 17 '18 at 16:50
• The probability of getting at least 4 is the same as getting exactly 4 OR exactly 5. You should know that in these kinds of situations "or" means addition and "and" means multiplication. Apr 17 '18 at 16:53
• But why do you multiply 35 time 3? Apr 17 '18 at 16:54
• Good question. Once you count the number of ways to choose 4 from 7, you still need to choose 1 from the remaining 3 in order to have chosen a total of 5.When getting exactly 4 correct out of 5: $7 \choose 4$ to account for the ones he gets correct, AND $3 \choose 1$ to account for the one he must get incorrect. Apr 17 '18 at 16:57
• @JimmySabater I incorporated some of the comments JungleShrimp made into my answer.
– John
Apr 17 '18 at 17:02