Could anyone explain me how to calculate the following...?

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Marko36

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Hi All,

I was never very good with math, but i have been breaking my brain trying to figure out how to calculate the odds of a positive outcome of the following scenario. This may be super easy but i just cant get my head around it.

So lets say there are 100 balls in a box, 50 black and 50 red. I need to randomly pick 10. How to calculate the odds of picking 10 same color balls?

Help here greatly appreciated. Thanks
 
Hi. Is this homework? Either way, what are your thoughts? Do you understand a more basic question about probability, like, what are the odds that the first ball picked is black, for example?

Please also familiarize yourself with the forum's guidelines. Thank you! :cool:
 
Marko36 said:
Hi All,

I was never very good with math, but i have been breaking my brain trying to figure out how to calculate the odds of a positive outcome of the following scenario. This may be super easy but i just cant get my head around it.

So lets say there are 100 balls in a box, 50 black and 50 red. I need to randomly pick 10. How to calculate the odds of picking 10 same color balls?

Help here greatly appreciated. Thanks
Click to expand...
Explaining probability theory is hard. It is not an intuitive idea.

\(\displaystyle \text {P(10 balls of same color in 10 picks)}\)

\(\displaystyle \text {P(10 red balls in 10 picks OR 10 blue balls in 10 picks) =}\)

\(\displaystyle \text {P(10 red in 10 picks) + P(10 blue in 10 picks) - P(10 red and 10 blue in 10 picks) =}\)

\(\displaystyle \text {P(10 red in 10 picks) + P(10 blue in 10 picks) - 0 =}\)

\(\displaystyle \text {P(10 red in 10 picks) + P(10 blue in 10 picks).}\)

But it should be intuitive that \(\displaystyle \text {P(10 red in 10 picks) = P(10 blue in 10 picks).}\)

\(\displaystyle \text {THUS, P(10 balls of same color in 10 picks) = 2 * P(10 red balls in 10 picks).}\)

\(\displaystyle 2 * \text {P(10 red balls in 10 picks) = }\)

\(\displaystyle 2 * \dfrac{50 * 49 * 48 * 47 * 46 * 45 * 44 * 43 * 42 * 41}{100 * 99 * 98 * 97 * 96 * 95 * 94 * 93 * 92 * 91} =\)

\(\displaystyle 2 * \dfrac{45 * 44 * 43 * 42 * 41}{2 * 99 * 2 * 97 * 2 * 95 * 2 * 93 * 2 * 91} =\)

\(\displaystyle 2 * \dfrac{(5 * 3 * 3) * (2 * 2 * 11) * 43 * (2 * 3 * 7) * 41}{2 * (3 * 3 * 11) * 2 * 97 * 2 * (5 * 19) * 2 * (3 * 31) * 93 * 2 * (7 * 13)} =\)

\(\displaystyle \dfrac{43 * 41}{2 * 13 * 19 * 31 * 97} \approx 0.119\%\)

It's about 12 chances in 10,000.
 
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JeffM said:
Explaining probability theory is hard. It is not an intuitive idea.

\(\displaystyle \text {P(10 balls of same color in 10 picks)}\)

\(\displaystyle \text {P(10 red balls in 10 picks OR 10 blue balls in 10 picks) =}\)

\(\displaystyle \text {P(10 red in 10 picks) + P(10 blue in 10 picks) - P(10 red and 10 blue in 10 picks) =}\)

\(\displaystyle \text {P(10 red in 10 picks) + P(10 blue in 10 picks) - 0 =}\)

\(\displaystyle \text {P(10 red in 10 picks) + P(10 blue in 10 picks).}\)

But it should be intuitive that \(\displaystyle \text {P(10 red in 10 picks) = P(10 blue in 10 picks).}\)

\(\displaystyle \text {THUS, P(10 balls of same color in 10 picks) = 2 * P(10 red balls in 10 picks).}\)

\(\displaystyle \text {P(10 red balls in 10 picks) = }\) … should have *2 attached

\(\displaystyle 2 * \dfrac{50 * 49 * 48 * 47 * 46 * 45 * 44 * 43 * 42 * 41}{100 * 99 * 98 * 97 * 96 * 95 * 94 * 93 * 92 * 91} =\)

\(\displaystyle 2 * \dfrac{45 * 44 * 43 * 42 * 41}{2 * 99 * 2 * 97 * 2 * 95 * 2 * 93 * 2 * 91} =\)

\(\displaystyle 2 * \dfrac{(5 * 3 * 3) * (2 * 2 * 11) * 43 * (2 * 3 * 7) * 41}{2 * (3 * 3 * 11) * 2 * 97 * 2 * (5 * 19) * 2 * (3 * 31) * 93 * 2 * (7 * 13)} =\)

\(\displaystyle \dfrac{43 * 41}{2 * 13 * 19 * 31 * 97} \approx 0.119\%\)

It's about 12 chances in 10,000.
Click to expand...
see red
 
Jeff, shouldn't that be ~.00119?

Can also be done this way:
(49! / 40!) / (99! / 90!) = .0011868....
 
Denis said:
Jeff, shouldn't that be ~.00119?
Click to expand...
6 of one or a half dozen of the other. \(\displaystyle 0.0019 = 0.119\%.\)

Can also be done this way:
(49! / 40!) / (99! / 90!) = .0011868....
Click to expand...
Indeed it can, and it is more efficient computationally. My personal feeing is that it is less intuitive. My experience has been that beginners find probability theory quite unintuitive so I try to stick with the more intuitive approach when possible. (I have great sympathy with those who struggle with probability: the first time I took a course in it, I dropped it because I could not grasp the way that professor approached the topic.)
 
JeffM said:
Explaining probability theory is hard. It is not an intuitive idea.

\(\displaystyle \text {P(10 balls of same color in 10 picks)}\)

\(\displaystyle \text {P(10 red balls in 10 picks OR 10 blue balls in 10 picks) =}\)

\(\displaystyle \text {P(10 red in 10 picks) + P(10 blue in 10 picks) - P(10 red and 10 blue in 10 picks) =}\)

\(\displaystyle \text {P(10 red in 10 picks) + P(10 blue in 10 picks) - 0 =}\)

\(\displaystyle \text {P(10 red in 10 picks) + P(10 blue in 10 picks).}\)

But it should be intuitive that \(\displaystyle \text {P(10 red in 10 picks) = P(10 blue in 10 picks).}\)

\(\displaystyle \text {THUS, P(10 balls of same color in 10 picks) = 2 * P(10 red balls in 10 picks).}\)

\(\displaystyle 2 * \text {P(10 red balls in 10 picks) = }\)

\(\displaystyle 2 * \dfrac{50 * 49 * 48 * 47 * 46 * 45 * 44 * 43 * 42 * 41}{100 * 99 * 98 * 97 * 96 * 95 * 94 * 93 * 92 * 91} =\)

\(\displaystyle 2 * \dfrac{45 * 44 * 43 * 42 * 41}{2 * 99 * 2 * 97 * 2 * 95 * 2 * 93 * 2 * 91} =\)

\(\displaystyle 2 * \dfrac{(5 * 3 * 3) * (2 * 2 * 11) * 43 * (2 * 3 * 7) * 41}{2 * (3 * 3 * 11) * 2 * 97 * 2 * (5 * 19) * 2 * (3 * 31) * 93 * 2 * (7 * 13)} =\)

\(\displaystyle \dfrac{43 * 41}{2 * 13 * 19 * 31 * 97} \approx 0.119\%\)

It's about 12 chances in 10,000.
Click to expand...

Thank you for your answer Jeff, this is very interesting. To be honest im still not quite getting it but i feel i'm getting closer. I'm actually surprised that the probability is that low, i would have thought its much higher.

Thanks again for your help, ill go and try to calculate the same with different numbers.
 
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