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- - **Coin and Dice problem**
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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, Lakshmi |

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? |

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} $ |

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

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