@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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