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Alice, Bob, and Carol repeatedly take turns tossing a die. Alice begins; Bob always follows Alice; Carol always follows Bob, and Alice always follows Carol. Find the probability that Carol will be the first one to toss a six. (The probability of obtaining a six on any toss is, independent of the outcome of any other toss.

 

This is how far I got:

 

A - B - C

 

Alice has a \(\frac{5}{6}\) chance of NOT rolling a \(6\) , and so does Bob. So if you go in the order of the game, there is a \(\frac{25}{36}\)chance that Alice and Bob won't roll a \(6\) during their turns. Since the next turn is Carol, the probability of her rolling a \(6\) is \(\frac{1}{6}\) . Therefore, the probability of Carol being the first one to toss a \(6\) is \(\frac{1}{6} * \frac{25}{36} = \boxed{\frac{25}{216}}\).

This is incorrect because the correct answer on the Art of Problem Solving website is \(\boxed{\frac{25}{91}}\)

Can someone please explain why I am incorrect?

 Feb 11, 2019
 #1
avatar+18280 
+1

Here is what might help:

https://math.stackexchange.com/questions/1327201/probability-that-on-three-rolls-of-dice-there-will-be-at-least-one-6-showing-up

 Feb 11, 2019
 #2
avatar+580 
+1

I understand now, and plus I discovered that its easier to just turn the question around and solve.

 Feb 11, 2019
edited by CalculatorUser  Feb 11, 2019
 #3
avatar+18280 
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Yah, it helped me too.....     cheeky

ElectricPavlov  Feb 11, 2019

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