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A coin with probability $p (0 < p < 1)$ of getting head, is tossed until a head appears for the first time. If the probability that the number of tosses required is even is $2/5$, then the value of $p$ is

  1. $2/7$
  2. $1/3$
  3. $5/7$
  4. $2/3$
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Given P(getting Head) = p which implies P(not getting head) = 1-p

Also given that number of tosses required are even for the head to appear for first time.

So,  getting first head on $2^{nd} $  toss = $(1-p)*(p)$ OR

getting first head on $4^{th}$ toss =  $(1-p)^{3}*(p)$ OR ....

Thererfore we obtain an infinite GP series i.e 

$(1-p)*(p)$ + $(1-p)^{3}*(p)$ + $(1-p)^{5}*(p)$+.............upto infinity = $2/5$

Solving above series we get $ p = 1/3$

3 votes
3 votes
Probability of heads = p

Success means heads at the even trial.

Let us see the first two trials. There are 4 possibilities :
e1. HH 2. HT 3. TH 4. TT

First two cases are failure. Third case is a success. In the fourth case, game continues and we end up at the same position as the beginning. Let $f$ be the probability of success which is equal to $2/5$

So the recurrence relation is :

$f = (1-p)* p + (1-p)^2 *f$

(probability of success = probability of success in second trial + probability of success in more than 2 trials )

Substituting $f = 2/5$ in the equation we get

$p = 1/3 $ as $p \neq 0$
There are other ways to solve this problem as well using infinite geometric progression. Others may give an alternate solution using that concept.
0 votes
0 votes

We need

  • tossed until a head appears for the first time. 
  • tosses required is even is $2/5$

So, here cases are

  • $TH$ with probability $\frac{1}{2^{2}}$
  • $TH$  with probability $\frac{1}{2^{4}}$
  • ...............

So, required probability will be $\frac{1}{2^{2}}+\frac{1}{2^{4}}+\frac{1}{2^{6}}+.........$

                                                   $=\frac{1}{2^{2}}\left (1+ \frac{1}{2^{2}}+\frac{1}{2^{4}}+\frac{1}{2^{6}}+....... \right )$

                                                   $\frac{1}{2^{2}}\left ( \frac{1}{1-2^{2}} \right )=\frac{1}{3}$

Now, check with equation $\left ( 1-p \right ).p+\left ( 1-p \right )^{3}.p+\left ( 1-p \right )^{5}.p+......$ where $p$ is probability of success.

In this equation put $p=1/3$, we will get $2/5$ as answer

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