Obviously 8 out of 20 or 40%.Originally Posted by rockwell
I don't quite understand. Do you mean "if at least one archer in a randomly chosen pair is left-handed"? In that case the solution is as follows:2. If a pair of archers is chosen at random is left-handed. Find the probability that exactly one of the pair is left-handed.
There are 20*19/2 = 190 possible pairs. Of these, 12*11/2 = 66 pairs consist only of right-handed archers, 8*7/2 = 28 consist only of left-handed archers, 190-66 = 124 have at least one left-handed archer, and 190 - 66 - 28 = 96 = 8*12 consist of a left-handed and a right-handed archer. So the probability is 96/124 = 24/31.
Let me know if this helps, then we'll get to the rest. Question 3 is done with the multiplication rule for conditional probabilities, 4 with total probability, 5 with Bayes rule.