a die is thrown 3 times . find the probability of -
i. getting three 1's
ii. getting three 1's given u have obtained 2 1's already
ans to i. is ofc 1/6^3
ans to ii. when derived using all conditional probability stuff comes 1/6
now i realise that P( 1,1,1 given 1,1 ) = P (  x y 1 ) where x and y are any random results we dont care about. i.e P( 1,1,1 given 1,1 ) is the same as probability of getting 1 on the 3rd try since 1,1 have already been achieved
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T
I
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A
