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When a coin is tossed, the probability of getting a Head is $p, 0 < p < 1$. Let $N$ be the random variable denoting the number of tosses till the first Head appears, including the toss where the Head appears. Assuming that successive tosses are independent, the expected value of $N$ is

1. $1/p$
2. $1/(1 - p)$
3. $1/p^2$
4. $1/(1 - p^2)$

$E = 1 \times p + 2 \times (1 - p)p + 3 \times (1 - p)(1 - p)p + \dots$

multiply both side with $(1 - p)$ and subtract:

$E - (1 - p)E = 1 \times p + (1 - p)p + (1 - p)(1 - p)p + \dots$

$= p /(1 - (1 -p)) = 1$  (because it is now forming a GP)

$=>(1 - 1 + p)E = 1$

$=> E = 1 / p$

So, Option (A)...
answered by Loyal (5k points) 8 53 81
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what this anwer means? atleast add some description about which step is taken why? what it implies..? anything? ?
It is an arithmetic geometric progression .That he is solving ...you can see how to solve it
@Vicky Bajoria, @Arjun sir

Is it possible for a random variable to have infinitely many values?

Actually, this is Geometric distribution.Here N is Geometric r.v  which represents no of tosses required for first head.

The expected value of a Geom r.v N = 1/P

So,

E(N) = 1/p

## The correct answer is (A)1/p

answered by Boss (8.9k points) 3 8 12