Question
There are three sections of a mathematics course available atconvenient times for a student. There is a 25% chance that Professor Jones gives a final exam,a 15% chance that Professor Cates givesa final exam, and a 9% chance thatProfessor Smithson gives a final exam. At other times there are twobiology sections, and in those the probabilities of a final are4% and 15%.Find the probability that a student who randomly chooses onemathematics course and one biology course has to take at least onefinal examination.
Explanation / Answer
There are 6 combinations of one math and one biology coursewhich we can list as follows (here m1, for example, means Jones'math course): {{m1,b1},{m1,b2},{m2,b1},{m2,b2},{m3,b1},{m3,b2}} Each of these course combinations has an equal chance of 1/6of being selected by the student. Replacing m1, m2, m3, b1, b2 withthe chance of NOT having a final (1 minus the numbers listed in theproblem) we can calculate the probability of not having any finalsas follows: 1/6({m1*b1}+{m1*b2}+{m2*b1}+{m2*b2}+{m3*b1}+{m3*b2}) Plug in the numbers and you get 45431/60000. Take one minusthis to find the probability of having one or more finals and youget 14569/60000 = 0.242817. So there is a little less than a 25%chance that the student will have to take at least one final.