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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. $1/(1 - p^2)$
asked in Probability by Veteran (18.3k points)   | 410 views

2 Answers

+11 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:

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

$  = 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 (4.7k points)  
selected by
what this anwer means? atleast add some description about which step is taken why? what it implies..? anything? ?
0 votes

N= P(1) + (1-P) (N+1)

Solving we will get N=1/P.

NOTE: 

N=expected number of tosses.

case1: Probability of getting head is p and require 1 toss i.e P(1=no. of tosses)

case2: if Tell appear means we have wasted one toss i.e. (1-P)(N +(1=one toss wasted))

 

 

answered ago by Veteran (21.6k points)  
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