# Probability

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## Probability

 You roll a fair 6 sided die, and then you flip a coin the no. of times shown by the dice. Find the expected value and the variance of the no. of heads obtained.
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## Re: Probability

 NB: please Dont rely on this solution until it is verified by Amit Sir. I have requested him to have a look at this. Let X be the number coming on the face of die. Let T be the number of heads when the coin is tossed x times. so f(t|x) = xCt (.5)^x  t=0,1, ... x E(T|X=x) = .5x , Var(T|X=x)= .25x E[T] = E[ E(T|X=x)  ] = (1/6) [ 1/2  + 2/2  + 3/2 .... 6/2] = 21/12 Var[T]= Var[   E(T|X)   ]    + E[    Var[T|X]  ]            Var[ E(T|X) ] = E[  ( E[T|X])^2  ]  - [E[  E(T|X) ] ]^2                           E[  ( E[T|X])^2  ] = 1/6[(1/2)^2 + (2/2)^2 .... (6/2)^2] = 91/24                           [E[  E(T|X) ] ]^2  =  [E[T] ]^2 = (21/12) ^2 =441/144            Var[ E(T|X) ] = 91/24 - 441/144 =105/144             E[  Var[T|X]  ] = 1/6[ Var[T|X=1] + ... Var[T|X=6]                                  =1/6 [1/4  +  2/4 ...6/4] = 21/24 Var[T]=105/144 + 21/24 = 231/144= 77/48 ---  "You don't have to believe in God, but you should believe in The Book." -Paul ErdÅ‘s
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## Re: Probability

 Administrator This post was updated on . In reply to this post by Devika You roll a fair 6 sided die, and then you flip a coin the no. of times shown by the dice. Find the expected value and the variance of the no. of heads obtained. Let N be the number of heads obtained. Let D be the number obtained on the dice. E(N) = E(E(N|D)) = E(D/2) = (1/2)E(D) = 7/4 V(N) = E(V(N|D)) + V(E(N|D)) E(V(N|D)) = E(D/4) = (1/4)E(D) = 7/8 V(E(N|D)) = V(D/2) = (1/4)V(D) = (1/4)(1/6) (2.5^2 + 1.5^2  + 0.5^2 + 0.5^2 + 1.5^2 + 2.5^2)                                                  = (1/4)(1/24) (25 + 9  + 1 + 1 + 9 + 25) = 70/96 Thus, V(N) = 154/96 = 77/48