Out of a pool of 50 questions, 7 random questions are selected for the exam. If a student prepares a total of 16 questions perfectly, what is the probability that he can solve at least one of the 7 random questions. My solution is :
- Number of possible question combinations:C(50,7)
- Combinations where no prepared question occurs:C(34,7)
P(no questions solved) : C(34,7)/C(50,7)
P(atleast one question) : 1 - (C(34,7)/C(50,7))
But I am not sure if this approach is correct. In addition, I have to also prepare a script with monte carlo method to verify it empirically. Which I am not sure how to proceed with.
Prob misses the first question is 34/50. Prob. missing second (after missing first) is 33/49, etc. so prob. missing all seven is $Q=(34\times 33\times ...\times 28)/(50\times 49 ...\times 44)$ The desired prob. is $P=1-Q$. Your answer for P(no questions) should be the reciprocal of your statement.