dse 2014 problem

[leon]

Am stuck with this one

There are two fair coins. Coin 1 is tossed 3 times. Let x be the no of heads that occur. After this coin 2 is tossed x times. Let y be the no of heads we get with coin 2. prob(x>=2|y=1) equals

[onionknight]

I'm getting 5/9 but it isn't there in the options and I can't find the answer key anywhere either

[Anuraag]

I think it is 1/2.

[onionknight]

How'd you get 1/2 ?

[Anuraag]

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[onionknight]

This doesn't look correct to me since Event B is 1 head on tossing coin X times and X could be 0,1,2 or 3.

[Dreyfus]

[leon]

thanks dreyfus. great help!

[onionknight]

Looks like all the options were incorrect then!

[L]

How did you calculate P(Y=1|X>= 2).
I did by intersection for numerator, P (X>=2 n Y=1)= P(X=2 n Y=1) + P(x=3 n Y=1) which I got correct.
But, if I do P(Y=1|X>= 2) (P(X=2) + P(X=3)), I am getting an extra 1/2.
I am doing it like this : P(Y=1|X>= 2) = (P(Y=1|X=2) P(X=2) +  P(Y=1|X=3)P(X=3)) (P(X=2) + P(X=3))