SOLUTION: How would I solve this problem: “A bag contains 4 red, 6 white and 9 blue marbles. How many ways can 5 marbles be drawn so that *at least* 2 marbles are blue?” I have tried

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Question 1202169: How would I solve this problem: “A bag contains 4 red, 6 white and 9 blue marbles. How many ways can 5 marbles be drawn so that *at least* 2 marbles are blue?”
I have tried the following but wonder if they one is correct:
1. 9C2 * 17C3 I thought that 2 blue marbles would be drawn, leaving 17 blue, red or white marbles to be drawn. The scenarios would be multiplied to yield 24480 ways.
2. Sum the results from each scenario of 2-5 blue marbles being drawn. (9C2 * 10C3) + (9C3 * 10C2) + (9C4 * 10C1) + (9C5) = 9486
Is there a flaw in my thinking?
Found 2 solutions by greenestamps, math_tutor2020:
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


I didn't check your calculations, but the second method is correct while the first is not.

Without trying to explain further, your first method won't give you the right answer because your "9" and "17" don't add up to 19.

Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

9C2*17C3 = 24480 is way too big compared to the upper bound 19C5 = 11628

11628 is the highest we can go.
This is one piece of evidence to allow us to rule out 9C2*17C3

The 2nd calculation you wrote is correct. The answer is 9486

9C2 * 10C3 = number of ways to get 2 blue and 3 non-blue
9C3 * 10C2 = number of ways to get 3 blue and 2 non-blue
9C4 * 10C1 = number of ways to get 4 blue and 1 non-blue
9C5 * 10C0 = number of ways to get 5 blue and 0 non-blue


Each row is of the form (mCn)*(pCq)
we have these properties
m+p = 9 blue + 10 nonblue = 19 marbles total
n+q = 5 selections allowed

Calculator verification using WolframAlpha
https://www.wolframalpha.com/input?i=%28%289choose2%29+*+%2810choose3%29%29+%2B+%28%289choose3%29+*+%2810choose2%29%29+%2B+%28%289choose4%29+*+%2810choose1%29%29+%2B+%28%289choose5%29%29
The input was "((9choose2) * (10choose3)) + ((9choose3) * (10choose2)) + ((9choose4) * (10choose1)) + ((9choose5))" without quotes.

The "choose" function is another name for the nCr combination function.

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Another approach:

There are 19C5 = 11628 ways to pick 5 marbles of any color. Order doesn't matter.

Also we have 10C5 = 252 ways to pick five non-blue marbles
And (9C1)*(10C4) = 9*210 = 1890 ways to pick one blue marble and four non-blue marbles.

Therefore, we have 11628 - (252+1890) = 9486 ways to select at least two blue marbles in a batch of five.

I'm using the idea that
n(0 blue) + n(1 blue) + n(at least 2 blue) = n(total)
which leads to
n(at least 2 blue) = n(total) - ( n(0 blue) + n(1 blue) )
n(at least 2 blue) = 19C5 - ( 10C5 + 9C1*10C4 )
n(at least 2 blue) = 11628 - ( 252 + 1890 )
n(at least 2 blue) = 9486

Calculator verification using WolframAlpha
https://www.wolframalpha.com/input?i=%2819+choose+5%29+-+%2810+choose+5%29+-+%289+choose+1%29*%2810+choose+4%29
The input was: (19 choose 5) - (10 choose 5) - (9 choose 1)*(10 choose 4)


Another relevant calculator
https://www.mathsisfun.com/combinatorics/combinations-permutations-calculator.html
It not only computes the nCr value, but it also lists out the various combinations in the "List Them" section.
I recommend exploring with small sets because listing tens of thousands of items would be a tedious task to verify.