Question 1202522: An urn contains 15 green, 10 blue, and 11 red balls. You take 3 balls out of the urn without replacement. What is the probability that all three balls are of the same color?
Found 2 solutions by Edwin McCravy, math_tutor2020:
Answer by Edwin McCravy(20060)   (Show Source):
You can put this solution on YOUR website!
P(all green) = C(15,3)/C(36,3) = ?/?
P(all blue) = C(10,3)/C(36,3) = ?/?
P(all red) = C(11,3)/C(36,3) = ?/?
Those are mutually exclusive so we may add their probabilities:
P(all the same color) = P(all green) + P(all blue) + P(all red) =
?/? + ?/? + ?/? = ?/?
Edwin
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
Answer: 37/357
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Work Shown:
A = probability of getting 3 green
B = probability of getting 3 blue
C = probability of getting 3 red
The ultimate goal is to compute A+B+C to get the probability of getting 3 of the same color.
Those events are mutually exclusive, which allows us to add the probabilities.
Let's calculate the value of A.
But first we need to find how many ways to get 3 green balls.
n = 15 green
r = 3 selections
The order doesn't matter so we use the nCr combination formula.
Pascal's Triangle is another alternative. But I'll use the formula.
n C r = (n!)/(r!(n-r)!)
15 C 3 = (15!)/(3!*(15-3)!)
15 C 3 = (15!)/(3!*12!)
15 C 3 = (15*14*13*12!)/(3!*12!)
15 C 3 = (15*14*13)/(3!)
15 C 3 = (15*14*13)/(3*2*1)
15 C 3 = 2730/6
15 C 3 = 455
There are 455 ways to get three green in a row without replacement.
Now we need to calculate how many ways there are to get any 3 balls regardless of color combos.
n = 15 green + 10 blue + 11 red = 36 total
r = 3 selections
n C r = (n!)/(r!(n-r)!)
36 C 3 = (36!)/(3!*(36-3)!)
36 C 3 = (36!)/(3!*33!)
36 C 3 = (36*35*34*33!)/(3!*33!)
36 C 3 = (36*35*34)/(3!)
36 C 3 = (36*35*34)/(3*2*1)
36 C 3 = 42840/6
36 C 3 = 7140
There are 7140 ways to select any three balls without replacement.
We therefore arrive at
A = 455/7140
that represents the probability of getting 3 green without replacement.
Through similar calculations you should find:
B = 120/7140
C = 165/7140
Those are the probability values for 3 blue and 3 red in that order.
Let me know if you need to see the steps.
Each fraction can be reduced, but I'll keep things un-reduced to easily add them in the next step.
Adding those mutually exclusive probability values will get us the final answer.
A+B+C = (455/7140)+(120/7140)+(165/7140)
A+B+C = (455+120+165)/7140
A+B+C = 740/7140
A+B+C = (20*37)/(20*357)
A+B+C = 37/357
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Another approach:
15 green + 10 blue + 11 red = 36 total
A = P(3 green) = (15/36)*(14/35)*(13/34) = 13/204
B = P(3 blue) = (10/36)*(9/35)*(8/34) = 2/119
C = P(3 red) = (11/36)*(10/35)*(9/34) = 11/476
Pay close attention to how the numerators and denominators count down by 1.
Eg: The numerators for B are 10,9,8. The denominators for any are 36,35,34.
A+B+C = 13/204 + 2/119 + 11/476
A+B+C = 91/1428 + 24/1428 + 33/1428
A+B+C = (91 + 24 + 33)/1428
A+B+C = 148/1428
A+B+C = (4*37)/(4*357)
A+B+C = 37/357
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