Assume you have a bag of 500 jelly beans. There are 40 different flavors. How many jelly beans must you randomly pick from the bag until the odds of drawing two of the same flavor is better than 50/50?
Assume you have a bag of 500 jelly beans. There are 40 different flavors. How many jelly beans must you randomly pick from the bag until the odds of drawing two of the same flavor is better than 50/50?
It depends on what color the jelly beans you pick are. If the first two you pick are the same color, you can stop there. If not, you need to keep going!
In a rush here, but I think...
p(n) = 1 x (1-1/39) x (1-2/39) x (1 - (n-1)/39)
Would be the probability of not having two of the same type. You would get to 50% at 8 beans I believe.
**sorry the probability of NOT having 2 of the same bean in that probability formula. 8 is still my answer, htough.
depends on factors wrote:
It depends on what color the jelly beans you pick are. If the first two you pick are the same color, you can stop there. If not, you need to keep going!
That doesn't answer the question.
Jesus, I did type not in the first one and I thought I missed it. I'm so wasted.
fisky wrote:
Assume you have a bag of 500 jelly beans. There are 40 different flavors. How many jelly beans must you randomly pick from the bag until the odds of drawing two of the same flavor is better than 50/50?
This is similar to the classic brainteaser about the minimum number of people you need to assemble in order to achieve better than 50/50 odds of two people sharing the same birthday (23).
fisky wrote:
Assume you have a bag of 500 jelly beans. There are 40 different flavors. How many jelly beans must you randomly pick from the bag until the odds of drawing two of the same flavor is better than 50/50?
And upon re-reading the original question, you would have to draw 7 beans for the odds of the NEXT bean (the 8th) to have a better than 50% chance of getting a flavor match.
So I guess the answer is 7.
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