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a) all throws are different

b) two throws are the same (a double)

c) three throws are the same (a treble)

d) four throws are the same (a quartet)

e) all five throws are the same (quintet).

So far i have worked out that:

a) (1/6)^5 . 6!/5!1! = 1/1296

b) 6 . (1/6)^2 . (5/6)^3 . 5!/3!2! = 1250/1296

c) 6 . (1/6)^3 . (5/6)^2 . 5!/2!3! = 250/1296

d) 6. (1/6)^4 . (5/6)^1 . 5!/4!1! = 25/1296

e) 6 . (1/6)^5 = 1/1296

I summated the above, expecting them to come to 1, but they come to 1527/1296

I then realised that i was double counting some of the outcomes because when a treble is thrown there is a chance a double is ALSO thrown and when a double is thrown there is a chance that a treble is thrown, so to deduct these i did

i) chance of throwing a double given that a treble is thrown AND not throwing a quintet=

5 . (1/6)^2 = 5/36 = 180/1296

ii) chance if throwing a treble given that a double is thrown AND not throwing a quintet=

5 . (1/6)^3 = 5/216 = 30/1296

1527/1296 - 180/1296 - 30/1296 = 1317/1296

Can anyone tell me why this still doesnt come to 1?