There are 50 red and 50 blue balls. Arrange them in 2 baskets, such that the probability of getting a blue ball is more.
Please help.
There are two baskets and we can put any number(even zero) balls in each of them.
I guess we can assume that x blue balls should be put in the first basket which has a total of n balls.
Then, we maximize
x/n +(50-x)/(100-n).
Is this correct? What is the answer?
Hello, kens!
There are 50 red and 50 blue balls.
Arrange them in 2 baskets, such that the probability of getting a blue ball is a maximum.
This is a classic problem/puzzle/riddle.
I'll hide the answer.
Drag your cursor over the region between the asterisks.
* Place one blue ball in one basket, the rest in the other basket.
The probability of getting a blue ball from Basket #1 is: 1
The probability of getting a blue ball from Basker #2 is: 49/99
P(Basket 1 and blue) = (1/2)(1) = 1/2
P(Basket 2 and blue) = (1/2)(49/99) = 49/198
Therefore: P(blue) = 1/2 + 49/198 = 148/198 ≈ 75% *