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Lakshmi77 July 21st, 2017 07:07 AM

Hello All,

Could you please help me in getting the solution for the below problem on probability?

If a man alternately tosses a coin and throws a die continuously, then find the probability of getting Head on the coin before he gets 4 on the die.

Thanks for your support,

Micrm@ss July 21st, 2017 08:50 AM

So let $X(n)$ be the event that the first head is on the n'th throw. Let $Y(n)$ be the event that the first $4$ is on the $n$'th throw.

Am I correct that you must find the probability $\mathbb{P}\{X<Y\}$. Now you can write this as follows:

$$\sum_{k=1}^{+\infty} \mathbb{P}\{X<k~\vert~Y = k\}\mathbb{P}\{Y=k\}$$

Do you agree with this? Does this help?

123qwerty July 21st, 2017 09:15 AM

We can re-word the problem a little bit:

Find the probability that he will get only the numbers 1, 2, 3, 5 or 6 before the first time he gets a head.

So the required probability is

$\displaystyle \frac{1}{2} + \left( \frac{1}{2} \right) \left( \frac{5}{6} \right) \left( \frac{1}{2} \right) + \left( \frac{1}{2} \right) \left( \frac{5}{6} \right) \left( \frac{1}{2} \right) \left( \frac{5}{6} \right) \left( \frac{1}{2} \right) + ...
= \frac{1/2}{1-5/12} = \frac{6}{7}$

Or we could do it in another way.

Find the probability that he will NOT get the number 4 before he gets a head for the first time.

$\displaystyle 1 - \left[ \left( \frac{1}{2} \right) \left( \frac{1}{6} \right) + \left( \frac{1}{2} \right) \left( \frac{5}{6} \right) \left( \frac{1}{2} \right) \left( \frac{1}{6} \right) + ... \right] = 1 - \frac{1/12}{1-5/12} = 1 - \frac{1}{7} = \frac{6}{7} $

Country Boy July 23rd, 2017 05:52 AM

It is impossible to flip a coin "continuously". You mean "continually".

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