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35 votes
35 votes

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. $\dfrac{1}{(1 - p^{2})}$
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K: Number of tosses till the first head appears

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

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

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


$Ans: A$

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Please explain how u did that ??
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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
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4 Answers

45 votes
45 votes
Best answer
$E = 1 \times p  + 2 \times (1 - p)p  + 3 \times (1 - p)(1 - p)p  + \dots$

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 $(A).$
edited by

4 Comments

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

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

probability of head =p

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

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

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
9 votes
9 votes
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
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7 votes
7 votes

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

P(Head) = p

So,

E(N) = 1/p

The correct answer is (A)1/p

2 Comments

I think this is the perfect answer
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edited by

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

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

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Answer:

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