# What is the probability that a boy who knows how to solve $25$ of potential $30$ questions will get at least $8$ of $10$ correct?

A boy is preparing for test. The teacher gives $30$ questions to study from and will select $10$ out of the $30$. The student only know hows to solve $25$ of the $30$ questions.

A)What is the probability that the student will get perfect on the test? B)what is the probability that the student will get at least $8$ questions correct?

I solved for A which is $11\%$, but I don't I have no clue how to solve B. Please give me a hint!

• – Fred
Jun 15, 2015 at 22:13

Note that $P(x \geq 8) = P(x=8) + P(x=9) + P(x=10).$

Here's a hint. I'll calculate the probability that the student gets exactly seven correct.

First, choose $7$ questions from the $25$ he knows ($_{25}C_7$). Then, choose $3$ from the $5$ he doesn't know ($_5C_3$).

Then the probability of getting a test that he knows exactly seven of them is

$$P(7) = \frac{_{25}C_7 \cdot _5C_3}{_{30}C_{10}}.$$

Can you take it from here?

If he get's at least 8 correct, that means he can get at most 2 wrong. Assuming he would never guess right (answer unknown question correctly), it makes more sense to figure out the probability that he will get 3 questions, 4 questions, or 5 that he doesn't want.

For example, the 4 questions he doesn't want on the test(Will only get 6 right): Choose 4 questions from the 5 he doesn't know : 5C4 Choose another 6 from the 20 he knows: 20C6 Then divide by the total ways of choosing 10 questions: 30C10. so: 5*38760/30045015 is about .6% ( I really hope I'm right here and not misleading you)

Could you get the 3 questions and 5 questions case then?

• You want to calculate the probability that the student gets at least eight questions correct, which is the sum of the probabilities that he gets eight, nine, or ten questions correct. Also, note that he knows how to answer $25$ of the $30$ questions that might appear on the test, not $20$. Jun 16, 2015 at 9:46