Probability
A bag contains n counters
One counter is blue and the rest are red.
Two counters are taken at random from the bag.
Express in terms of n, the probability that a counter of each colour is taken. Give your answer as a fraction in it's simplest form.
One counter is blue and the rest are red.
Two counters are taken at random from the bag.
Express in terms of n, the probability that a counter of each colour is taken. Give your answer as a fraction in it's simplest form.
Original post by DrDsy
Probability of blue: 1/n
Probability of red: n-1/n
Probability of getting one of each = (1/n) * (n-1/n) = n-1 / n^2
I think this is correct
Probability of red: n-1/n
Probability of getting one of each = (1/n) * (n-1/n) = n-1 / n^2
I think this is correct
Not quite.
If the first one is blue, think more carefully about the second choice.
Also don't forget that (red, blue) would also be valid.
However, try and get the OP to work through the solution, rather than putting all the detail in.
(edited 5 years ago)
I would use combinatorics which would be solved as {(1C1)[(n-1)C1]}/(nC2)
which you can resolve as [1*(n-1)]/{[n(n-1)]/2}
then as 2*(n-1)/[n(n-1)]
cancelling out the (n-1) will produce 2/n
or else you can solve by making all conditions into account
(blue,red): (1/n)(n-1/n-1) = 1/n
(red,blue): (n-1/n)(1/n-1) = 1/n
(blue,red)+(red,blue) = 1/n + 1/n = 2/n
which you can resolve as [1*(n-1)]/{[n(n-1)]/2}
then as 2*(n-1)/[n(n-1)]
cancelling out the (n-1) will produce 2/n
or else you can solve by making all conditions into account
(blue,red): (1/n)(n-1/n-1) = 1/n
(red,blue): (n-1/n)(1/n-1) = 1/n
(blue,red)+(red,blue) = 1/n + 1/n = 2/n
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