- #1
battery2004
- 9
- 0
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[tex]^{3}_{60}[/tex]
The favorable cases would be - C[tex]^{3}_{40}[/tex]
So the answer is C[tex]^{3}_{40}[/tex] / C[tex]^{3}_{60}[/tex]
b)
Same as a) but the favorable cases are C[tex]^{2}_{40}[/tex]
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.
Last edited: