I have 100 candies in a jar. 90 are lemon and 10 are mint. every day I take a candy out. If it is lemon, I eat it, but if it is mint, I put it back and take out another candy. If the new candy is also mint, then I eat it. What is the probability that the last candy I eat is a mint?

Warning: Use of undefined constant bfa_comments - assumed 'bfa_comments' (this will throw an Error in a future version of PHP) in/home/customer/www/riddledude.com/public_html/wp-content/themes/atahualpa/comments.phpon line132WOW! The chances of the 1st candy you eat being mint is 10/100 * 9/99 = 0.90909…%

In order to eat a mint candy, there will always be a mint in the jar – so you can never eat every candy. Eventually there may be just one mint candy left in the bowl. At that time, you pick the mint candy, put it back, but then are unable to take out “another” candy – because there will not be another one in there. But the day before you could have eaten either a mint or a lemon – I think the math is too hard for me to figure out? Is there a trick?

Math is never that tricky, just not that fun.

The probability of eating the mint candy = probability of drawing 2 consecutive mint candies = 90c2 / 100c2 = 90*89 / 100*99 = 89/110

The probability of eating a lemon candy = probability of drawing a lemon candy = 1/10

Thus, the probability of the last candy eaten being mint = P(M) / [P(M) + P(L)] = 89/100

Let’s call this one solved and not hurt our brains any more this early in the morning.