mrk9414 wrote:
SherpaPrep wrote:
Since there is only one yellow ball, of course neither the 1st nor the 2nd can be yellow. But there are two scenarios which can result in the 2nd ball being "not red."
In scenario 1, the 1st ball chosen is red, and the 2nd ball is either blue or green. The odds of three events happening should multiplied. So, the odds of the 1st being red are 2/5, the 2nd being either blue or green are 2/4, and the odds of the 3rd being yellow are 1/3:
2/5 x 2/4 x 1/3 = 1/15
In scenario 2, the 1st ball chosen is either blue or green and the 2nd ball is whichever out of blue or green didn't get chosen the first time. So, the odds of the 1st being either blue or green are 2/5, the 2nd being the other one are 1/4, and the odds of the 3rd being yellow are 1/3:
2/5 x 1/4 x 1/3 = 1/30
Adding these two probabilities, 1/15 and 1/30, results in an answer of 1/10.
Hey there,
I am really confused about when to add or multiply the probabilities?
I am trying to follow this rule but the step seems different
If all the events happen (an "and question") Multiply the probabilities together.
If only one of the events happens (an "or question") Add the probabilities together
would you shed some light on it?
Hi There!
Refer the below:
**And means that the outcome has to satisfy both conditions at the same time.- Multiply
**Or means that the outcome has to satisfy one condition, or the other condition, or both at the same time.- Add
You can also refer the above book for detailed explanation:
Maths book-
https://greprepclub.com/forum/gre-math- ... -2609.htmlTo explain the above question:
We know the third ball is yellow thus, the probability to select a yellow ball= 1/5
The second ball is not red, which means it is either blue or green= 2/4= 1/2
And the first ball is either red or green or blue depending on the second ball= 3/3=1
Now, using the above rule
This question uses the And rule:\(\frac{ 1}{5 }\)x \(\frac{1}{2}\)
=\(\frac{1}{10}\)
Answer:\(\frac{1}{10}\)
Hope this helps!