Four men, A, B, C, and D, line up in a row. What is the probability that man A is at either end of
the row?
(from barron’s)
The answer is 1/2
Can someone explain why? (I think it should be 1/12
I thought so too, its probably an error
Man A will end up in one of the four positions. Each position has equal probability. Since there are 2 ends to the row the chances he ends up on the end are 2 of the 4 possibilities or 2/4 or 1/2.
The total possible arrangements will be 4! = 24. The first man can be at either the first or the last place in 2 ways. So the probability is 2/24 = 1/12
No, you’re solving a different problem. In the problem discussed here, we are simply focused on one person who can be in any of 4 places. We ant to know the odds of his being in either of two out of those 4 places. The answer is 1/2 as stated in the OP.
It is 1/2 indeed. If you fix A at the front, there are 3! = 6 ways to arrange B, C, D. Same with A in the back. 12/24 = 1/2
@MITer94 - to me, that calculation overcomplicates things resulting in the potential for a computation error. Since three of the people (B,C and D) are irrelevant to the question, why not just focus on the possible position for A? You have more experience than me here though, which method do you think works better for the “typical” student?
@CHD2013 I know exactly what you mean; my intent was as an “alternate” solution. Sometimes I like to just directly compute the number of ways if it isn’t too computational. Plus, it may becomes clear to some that exactly 12 ways work (out of 24).