@kunal patil:
U were proceeding in correct way...in next series u cud hav seen a
formation of arithmatico geometric series...
It doesnt matter what the value of no of faces in a dice is..ans will
be always 2...:)
My simplified soln: o+e+1
o=probability of odd number coming in 1
throw of dice
In case o & e are mutually exclusive the
ans is always 2...(and in even odd case like
here e+o=1...so ans is always 2...let n be
anything)
On 8/8/11, Kunal Patil <kp101...@gmail.com> wrote:
> @Shady :
> No...we can say this only at the time when following constraints are
> satisfied:
> 1) *Outcome* of event *should be* *binary*. (In above example Sum can have
> binary outcomes only i.e. EVEN or ODD)
>
> 2) Random variable x in P(x) should be supported on set {1,2,3,4,....} i.e.
> It *should start from 1* and then take on *contiguous values*.
>
> 3) *Sequence* *of probabilities* for x,x+1,x+2,... *must form a GP*.
> (In above sum P(1)=1/2; P(2)=1/4; P(3)=1/8 forms GP)
>
> Taking another simple example having biased coin:
> P(H) --> 1/4
> P(T) --> 1 - 1/4 = 3/4.
> Say we want to find out expected number of coin tosses required to get first
> head.
> (Outcome is binary and random variable starts from 1 & can take contiguous
> values.Also, sequence of probabilities form a GP.)
>
> You may verify that answer comes out to be 1/P(H) --> 1/(1/4) --> 4
>
> @ never_smile....: If I understand your question correctly then Probability
> of getting even sum in one roll is P(2) + P(4)....
> For e.g. say
> P(1) --> 1/5
> P(2) --> 2/5
> P(3) --> 1/5
> P(4) --> 0
> P(5) --> 1/5
>
> P(even in one roll) --> 2/5 + 0 --> 2/5.
>
> P(even in one roll) = 2/5;
> P(even in 2 rolls) = (3/5)*(3/5);
> P(even in 3 rolls)=(3/5)*(2/5)*(3/5)
> This probability sequence doesn't form a GP.
> Thus, above formula should not be applied & you should calculate E[x] by
> trivial method.
>
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