Temurkhon wrote:
sharmasneha wrote:
Each of 4 bags contains 25 blue disks, 25 green disks, 25 orange disks, 25 yellow disks, and nothing else. If one disk is chosen at random from each of the four bags, what is the probability that the number of blue disks chosen will be no less than 1 andno greater than 3?
(A) 1/16
(B) 41/128
(C) 87/128
(D) 225/256
(E) 255/256
I have realized that we should multiply all non scenarios because problem says no less AND no more. If it said OR we could add and get above stated answers. My own speculation it that, because we are saying about undesirable conditions, we should add to get higher probability and subtract it from 1. Could experts explain, please
Dear
Temurkhon I understand that what confused you is a bit tricky word - AND - used in the question. Here's my take on it.
what is the probability that the number of blue disks chosen will be no less than 1
andno greater than 3
The above statement means that we are looking for 1, 2 and 3. If we do 1-(P0 * P4) we are entering an impossible territory. It is not possible to have both the conditions at the same time i.e. We can't have P0 and P4 at the same time.
Lets understand it this way :
We have the following probabilities for blue :
P0
P1
P2
P3
P4
and at a time, we can have only one of these, i.e. they are mutually exclusive.
SO, to eliminate the possibility of P0 AND P4, we need to add them and then eliminate the chances of their occurrence.
Hope that helps you.