Probability

[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.

[Sinistral]

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

[Amit Goyal]

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



                   

[Devika]

Amit Sir,
Is there a way to go about this problem without the use of conditional probabilty?

[Amit Goyal]

Yes, There are many ways to do this problem. But I gave you the way that is computationally least cumbersome.