On Halloween night, a young boy was trick-or-treating. It was late at night and he was on his last stop. He knocked at the door and said:
Trick-or-Treat!
The man inside said he had a bag of just 4 candy left. He said there were 1 hard candy, 1 gummy, and 2 lollipops.
He allows him to take 2 candy, and the boy does so. He grabs 2 from the bag without looking! He then states he has at least 1 lollipop, afterwards he asked his mom what the chances are of him having 2 lollipops.
His mom had no idea, but I bet you do! What were the chances of the boy pulling out 2 lollipops that night?
Answer
The answer is
1 in 5.
because given that he has at least 1 lollipop, the possibilities are:
1 lollipop, 1 hard candy (probability 1 in 8)
1 lollipop, 1 gummy (probability 1 in 8)
2 lollipops (probability 1 in 16)
So the conditional probability of him having 2 lollipops is $\frac{1/16}{\frac{1}{8}+\frac{1}{8}+\frac{1}{16}}=\frac{1}{5}.$
At the OP's behest, I'm also explaining why it looks like the answer should be 1 in 3. (If anyone's not confused already, let's confuse 'em! :-) ) The boy has 1 lollipop; there are 3 possibilities for the other sweet, of which only 1 is 'the other lollipop'; so it looks like the answer should be 1 in 3. But we know nothing to distinguish the two lollipops. He has 1, yes, but which one did we choose? If the lollipops were one red and one yellow, and we knew he had the red one, then the answer would be 1 in 3. As it stands, however, the answer is 1 in 5 as proved above.
No comments:
Post a Comment