Answer:
39/812
Step-by-step explanation:
The total no. of balls is 6+7+8+9 = 30.
There are 4 possible outcomes that fits "all 3 balls taken are of the same color": 1. all red balls/2.all green balls / 3. all blue balls/4. all white balls. Either one will do. So the required probability is the sum of probability that these 4 outcomes happen.
First lets find the probability of taking out all red balls.
There are total 30 balls, but only 6 are red, so in the first try, the probability of taking out a red ball is 6/30.
In the 2nd time, since you have already taken out one red ball, the no. of red balls left is 5, and the total no. of balls is 29. So the probability of taking out another red ball is 5/29.
This repeats for the 3rd time. The probability of taking out the 3rd red ball is 4/28.
These 3 times all have to happen in order, so the total probability is
6/30 x 5/29 x 4/28
=1/203
Now find the probability of taking out all green balls. The process is same from finding the probability of taking out all red balls, just that replace 6 to 7 instead.
So, the probability of taking out all green balls is
7/30 x 6/29 x 5/28
=1/116
Repeat again for blue balls and white balls.
Probability of taking out all blue balls = 8/30 x 7/29 x 6/28 = 2/145
Probability of taking out all white balls= 9/30 x 8/29 x7/28 =3/145
Now add all 4 fractions up.
1/203 + 1/116 + 2/145 + 3/145
=39/812
By the way, another method is to use combination "nCr". You will also get the same answer.
n: number of items
r: number of items being chosen at a time
(Probability of all red balls + Probability of all green balls + probability of all blue balls + probability of all white balls)÷Total no. of combinations - which is random 3 balls out of 30.
(6C3 + 7C3 + 8C3 +9C3) / (30C3)
= 39/812
