Math Is Fun Forum

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Hi, here's my two cents but do take it with a grain of salt!
My notation: denote C1 as first match Canada vs Canada; M2 as second match Mexico vs Mexico; CM3 as third match Canada vs Mexico; and so on...

Dealing with the first sub-problem: 'only 1 pairing comprised of 2 Canadian team'
One such arrangement is C1,CM2,CM3,...,CM10.
Let's compute the probability of this arrangement:
P(C1,CM2,CM3,...,CM10) = (11/20*10/19)*(9/18*9/17*2)*(8/16*8/15*2)*(7/14*7/13*2)*...*(2/4*2/3*2)*(1/2*1/1*2) = 0.003048...
This is only one possible arrangement, another arrangement is CM1,C2,CM3,CM4,...,CM10 for example.
There are 10 of these arrangements, therefore the answer to the first subproblem is 0.03048.

So one possible strategy for the first 8 subproblems can be summarised as:
1. Compute the probability of one such arrangement satisfying the conditions
2. Find the total number of arrangements
3. Multiply them both for the overall probability (you can do this because the probability of each arrangement should be the same).

The solution to the last two subproblems is different.
Hope this helps and good luck!

Hi ganesh!
Just wanted to pop by after my many years of absence and great to see you still here! Hope all is well.

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Heya ganesh!

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