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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. $\dfrac{1}{p}$
2. $\dfrac{1}{(1 - p)}$
3. $\dfrac{1}{p^{2}}$
4. $\dfrac{1}{(1 - p^{2})}$

K: Number of tosses till the first head appears

$K=p\times 1+ (1-p)\times (K+1)$

$K=p+K+1-Kp-p$

$K=\dfrac{1}{p}$

$Ans: A$

Please explain how u did that ??
1.If we get a tail in the first throw that means we have wasted one throw..ie (probability of getting tail)(N+1)= (1-p)(N+1)

2.If we get a head on the first throw than  we are done.. ie (Probability of getting heads)(No. of tosses)=(p)(1)=p

Summing both up..we get.. N=(1-p)(N+1)+ p

After solving we get N=1/p

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

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

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

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

$\quad \implies (1 - 1 + p)E = 1$

$\quad \implies E = 1 / p$

So, option $(A).$

Why did you multiply each term with successive numbers like 1,2,3........??
edited

N be the random variable denoting the number of tosses till the first Head appears,

## probability of tail=1-p

Toss1 "tail" * toss2 "tail"*............upto N-1 times * head appears only one time in last

{(1-p)^N-1]  *  p

is it right??????????

### if yes   then how can solve after (1-p)^N-1 * p

N can be 1,2,3, ….
So we are just adding them

Let X be the number of tosses of a fair coin until a head appears, and we want to find E(X). When we toss the coin once, there are two possibilities:

• a head occurs with a probability p.
• a tail occurs with a probability (1-p). In this case the expected number of trials would be (1 + E(X)) where 1 is from the lost trial

∴ E(X) = p*1 + (1-p)*(1+E(X))
Solving it gives us E(X)=1/p.

by
 Tosses Probability 1 p 2 (1-p)p 3 (1-p)(1-p)(p) 4 (1-p)(1-p)(1-p)p . . . . . . . . n $(1-p)^{n-1} \times p$

$E = 1 \times p + 2 \times p(1-p) + 3 \times (1-p)^{2}p + .........$

$E = \Large \sum_{n=1}^{\infty} n \times (1-p)^{n-1} \times p$

$E = \Large p \sum_{n=1}^{\infty} n \times (1-p)^{n-1}$

$E = \Large p \sum_{n=0}^{\infty} (n+1) \times (1-p)^{n}$

and we know that,

$\Large \sum_{k=0}^{\infty} (k+1)x^k = 1 + 2x + 3x^2 + 4x^3+..... = \frac{1}{(1-x)^2}$

Hence,

$E = \frac{1}{(1-(1-p))^2} \times p$

$E = \frac{1}{(p)^2} \times p$

$E \large = \frac{1}{p}$

### 1 comment

I think better way to solve

T T T T H

P(H)=1/5

here N=5

T T T T T T T H

P(H)=1/7

here N=7

so N=1/P

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

by

I think this is the perfect answer
edited

For reference, one can see what is expected value of geometric random variable here:

Expected Value of Geometric Random Variable..Proof @ Khan Academy