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

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 $(A).$
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29 votes
29 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.

12 votes
12 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,

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

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

$E  \large = \frac{1}{p}$

 

 

8 votes
8 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

Answer:

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