Lost with Probability homework

[battery2004]

Homework Statement

A student has learned answers to 40 questions out of 60. A student has to pick 3 questions. What is the probability, that the student will know:
a) all 3 questions
b) exactly 2 questions
c) at least 2 questions

The Attempt at a Solution

a)
So all the possible cases would be - C$$^{3}_{60}$$

The favorable cases would be - C$$^{3}_{40}$$

So the answer is C$$^{3}_{40}$$ / C$$^{3}_{60}$$

b)

Same as a) but the favorable cases are C$$^{2}_{40}$$

c)

So the probability for one question to be right would be 40/60 = 2/3.

Probability that all the questions are right (2/3)^3
Probability that 2 of the questions are right and one is wrong (2/3)^3 * (1/3)

so

(2/3)^3 + 2*((2/3)^2*(1/3)) = 16/27

Can someone confirm these answers?

Thanks in advance.

 

[vela]

Your answer to (a) is correct, but your answers to (b) and (c) aren't.

For (b), you calculated the number of ways you can choose two correct answers, but you actually want the number of ways to choose two correct answers and one wrong answer.

For (c), you can't assume the probability remains at 2/3 for each pick. Once the student has answered a question, he or she has a smaller pool of questions to choose from. In other words, there's no replacement after each trial.

The idea behind part (c) is that you can find its answer in terms of the answers to (a) and (b).

 

[battery2004]

Thanks for you answer.