I came across an interesting question when i was teaching probability to my juniors:
An assignment consisting of 20 questions is given in the class. It is mentioned that 8 among these willform the upcoming class test. You have learned the answered to 12 questions whereas your friend isable to learn only 10 questions. What is the probability that you friend scores strictly higher than you?Assume that there are no part marks and a question either gets a full mark or none.
i tried to solve thus by considering an urn consisting of 20 marbles. Suppose if i mark 12 of them (with red dot) and put them back in if i ask someone to pick any 8 from the urn uniformly independent of the markings. So let X denote the number of marked ones with red dot and Y denote the number of marked ones with blue dot. Then X corresponds to my marks and Y corresponds to my junior'smarks.Now i am stuck at Objective to find the probabilityP(Y > X).
How will i find it and what will be the answere?
I think we are supposed to assume that your 10 questions, and your friend's 12 questions, were chosen at random from the set of 20 questions on the test, and your choices are independent of your friends'.
The probability that you can answer exactly $x$ questions out of the $8$ questions on the test is $$f_X(x) = \frac{\binom{10}{x}\binom{10}{8-x}}{\binom{20}{8}}$$ for $0 \le x \le 8$, and the probability that your friend can answer exactly $y$ questions is $$f_Y(y) = \frac{\binom{12}{y}\binom{8}{8-y}}{\binom{20}{8}}$$ for $0 \le y \le 8$.
So the probability that your friend will answer more questions than you is $$P(Y>X) = \sum_{y=1}^8 \sum_{x = 0}^{y-1} f_X(x) f_Y(y)$$