My Math Forum Coin and Dice problem

 July 21st, 2017, 07:07 AM #1 Newbie   Joined: Jul 2017 From: Hyderabad Posts: 1 Thanks: 0 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 Last edited by skipjack; July 21st, 2017 at 08:59 AM.
 July 21st, 2017, 08:50 AM #2 Senior Member   Joined: Oct 2009 Posts: 733 Thanks: 247 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 $\mathbb{P}\{X  July 21st, 2017, 09:15 AM #3 Senior Member Joined: Dec 2012 From: Hong Kong Posts: 853 Thanks: 311 Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics We can re-word the problem a little bit: Find the probability that he will get only the numbers 1, 2, 3, 5 or 6 before the first time he gets a head. So the required probability is$\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} \$ Thanks from Lakshmi77
 July 23rd, 2017, 05:52 AM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 896 It is impossible to flip a coin "continuously". You mean "continually".

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