himadribora wrote:I am not too sure, but this is how it should be:
P(odd) = P (even) = 1/2 (because there are 50 odd and 50 even numbers)
Sum or the three numbers can be odd only under the following 4 scenarios:
Odd + Odd + Odd = 1/2 x 1/2 x 1/2 = 1/8
Odd + Even + Even = 1/2 x 1/2 x 1/2 = 1/8
Even + Odd + Even = 1/2 x 1/2 x 1/2 = 1/8
Even + Even + Odd = 1/2 x 1/2 x 1/2 = 1/8
Other combinations of odd and even will give even numbers.
Adding up the 4 scenarios above= 1/8 + 1/8 + 1/8 + 1/8 = 4/8 = 1/2. (P.S: You will have to add and not multiply)
DennaMueller wrote:I wrote out all of the possible picks:
4 of the possible picks are even and 4 are odd, so 4/8 or 1/2
mschwrtz wrote:Nice work all. Here's a very brief way to solve:
The sum of the first two balls will be either even or odd.
If the sum of the first two balls is even, then there is a probability of 1/2 that the sum of the three balls will be odd (because exactly half the balls available for the third draw are odd).
If the sum of the first two balls is odd, then there is a probability of 1/2 that the sum of the three balls will be odd (because exactly half the balls available for the third draw are even).
So whatever the sum of the first two balls, the probability that the third will yield an odd sum is 1/2.
tony.duran wrote:Is it possible to solve this problem with the anagram method?
herogmat wrote:Hi , nice solution. I was just thinking if the problem told us to calculate the probablity for the case of 'without replacement' what would have been the easy way to do that. I mean if the ball was taken out from the container after selection then how should we proceed ?
michael_shaunn wrote:I have a question if anyone can plz help.
Why four cases instead of 2?
Why O+E+E,E+O+E and E+E+O instead of just one of the three?
"Three balls are selected at random"..Does this mean that the balls have been selected three at a time or three balls one by one?
As per me..."three balls are selected at random" means that the three balls are selected all at once instead of one by one.