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11-01-2012, 01:24 PM
(This post was last modified: 09-01-2022, 01:01 AM by Fearless Community.
Edit Reason: Checked Encoding
)
Well I thought I would have a try at this, no idea if I got it correct but I am fairly sure I am there.
Firstly, I organised all the different combinations of possibilities that occur with only 2 of the 5 being red marbles. This I completed with a probability tree.
I got:
1. p(R,R,W,W,W)
2. p(R,W,R,W,W,)
3. p(R,W,W,R,W)
4. p(R,W,W,W,R)
5. p(W,R,R,W,W)
6. p(W,R,W,R,W)
7. p(W,R,W,W,R)
8. p(W,W,R,R,W)
9. p(W,W,R,W,R)
10. p(W,W,W,R,R,)
Now after this, I calculated the overall probabilites of each of these occurences.
For example, No. 1 would look like this:
3/10 x 2/9 x 7/8 x 6/7 x 5/6 = 1/24.
Interestingly, each of these come to a total of 1/24 and so in the end you get 10/24 which of course simplifies to 5/12.
I figure this may be the answer as it is a nice, even fraction that looks right - typical of school given questions.
My final answer is 5/12.
Firstly, I organised all the different combinations of possibilities that occur with only 2 of the 5 being red marbles. This I completed with a probability tree.
I got:
1. p(R,R,W,W,W)
2. p(R,W,R,W,W,)
3. p(R,W,W,R,W)
4. p(R,W,W,W,R)
5. p(W,R,R,W,W)
6. p(W,R,W,R,W)
7. p(W,R,W,W,R)
8. p(W,W,R,R,W)
9. p(W,W,R,W,R)
10. p(W,W,W,R,R,)
Now after this, I calculated the overall probabilites of each of these occurences.
For example, No. 1 would look like this:
3/10 x 2/9 x 7/8 x 6/7 x 5/6 = 1/24.
Interestingly, each of these come to a total of 1/24 and so in the end you get 10/24 which of course simplifies to 5/12.
I figure this may be the answer as it is a nice, even fraction that looks right - typical of school given questions.
My final answer is 5/12.