# Probability of Getting at least 2 correct answers out of 7 (3 choices are correct)?

Imagine there's a multiple choice question with 7 possible choices. 3 are correct and a student randomly selects 3 choices. What's the probability that he gets at least 2 correct?

I thought it was:

P(exactly 2 right) + P(3 right) = (3C2 * 4C1) / 7C3 + (1 / 7C3)

Is this correct?

Thanks, Mariogs

How many ways can we have at least two correct (equiv at least 2 red balls), is either having exactly three correct or two correct and one wrong. As you wrote, you have $\binom{3}{2}\cdot\binom{4}{1}$ number of ways to get exactly two correct (which two correct times which one wrong), and you have $\binom{3}{3}\cdot\binom{4}{0}$ number of ways to get all three correct.
Finally, the number of total possible situations is $\binom{7}{3}$
So, combining this information: $\dfrac{\binom{3}{2}\cdot\binom{4}{1} + \binom{3}{3}\cdot\binom{4}{0}}{\binom{7}{3}}$ is the probability.
Long story short, yes your answer was correct, noting that fractions can be combined and that $\binom{3}{3}\cdot\binom{4}{0}=1$