+0  
 
0
37
2
avatar+176 

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+3535 
+2

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

 

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

.
 Jan 3, 2019
 #2
avatar+176 
+1

Thank you so much!

 Jan 3, 2019

10 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.