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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

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As I understand the question, there is a single problem on a multiple choice test that we are interested in, and the problem asks for three responses total out of 7 possible. The student wont pick the same answer multiple times in a row so we can think of this as a probability problem "without replacement". An equivalent situation is that he has a bucket of seven balls, three of which are red and four of which are white, after picking three balls randomly without replacement what the chance is that he picks at least two red balls.

Discrete probability like this says the total number of outcomes with the property that we are interested in divided by the total number of outcomes regardless of satisfying our property.

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 $\binom{3}{3}\cdot\binom{4}{0}=1$

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