2
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$
    – Peter
    Apr 17 '18 at 16:41
  • $\begingroup$ 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. $\endgroup$ Apr 17 '18 at 16:42
2
$\begingroup$

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 $$P_5 + P_4 = 21/252 + 105/252 = 126/252 = 0.5.$$

$\endgroup$
5
  • $\begingroup$ it is unclear how you get the second probaility. why you adding? $\endgroup$
    – James
    Apr 17 '18 at 16:50
  • 1
    $\begingroup$ 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. $\endgroup$ Apr 17 '18 at 16:53
  • $\begingroup$ But why do you multiply 35 time 3? $\endgroup$
    – James
    Apr 17 '18 at 16:54
  • 1
    $\begingroup$ 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. $\endgroup$ Apr 17 '18 at 16:57
  • $\begingroup$ @JimmySabater I incorporated some of the comments JungleShrimp made into my answer. $\endgroup$
    – John
    Apr 17 '18 at 17:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.