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A turn consists of rolling a standard die and tossing a fair coin. The game is won when the die shows a 1 or a 6 and the coin shows heads. What is the probability the game will be won before the fourth turn? Express your answer as a common fraction.

 Jan 3, 2019
 #1
avatar+4424 
+2

\(P[\text{win 1 game}] = \dfrac 1 3 \dfrac 1 2 = \dfrac 1 6\\ P[\text{win before 4th turn}] = P[\text{win game 1}]+P[\text{lose 1, win 2}]+P[\text{lose 1,2, win 3}]\)

 

\(P[\text{win game 1}] = \dfrac 1 6\\ P[\text{lose 1, win 2}] = \dfrac 5 6 \dfrac 1 6 = \dfrac{5}{36}\\ P[\text{lose 1,2, win 3}] = \left(\dfrac 5 6\right)^2 \dfrac 1 6 = \dfrac{25}{216}\\ P[\text{win before 4th turn}] = \dfrac 1 6 + \dfrac{5}{36} + \dfrac{25}{216} = \dfrac{91}{216}\)

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 Jan 3, 2019
 #2
avatar+183 
+1

Thank you so much!

 Jan 3, 2019

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