Question

A red die is rolled a single time. A green die is rolled repeat- edly. The game stops the first time that the sum of the red and green die is either 4 or 7. What is the probability that the game stops with a sum of 4?

Answer 1

Answer 2

Calculation:

1. Red Die Outcomes: The red die shows a number from 1 to 6, each with probability 1/6.

2. 
   

3. Stopping Conditions: The game stops when the sum of the red and green die is either 4 or 7.

4. 
   

5. Possible Sums for 4 and 7:

• Sum = 4: (Red=1, Green=3), (Red=2, Green=2), (Red=3, Green=1).

• Sum = 7: (Red=1, Green=6), (Red=2, Green=5), (Red=3, Green=4), (Red=4, Green=3), (Red=5, Green=2), (Red=6, Green=1).

• 
  

6. Probabilities:

• Green=3 (sum=4) or Green=6 (sum=7).

• Probability: (1/6) * (1/6 + 1/6) = (1/6) * (2/6) = 2/36.

• For each red die outcome, calculate the probability that the green die results in a sum of 4 or 7.

• For example, if Red=1:

• Repeat for all red die outcomes and sum the probabilities for sums of 4 and 7 separately.

• 
  

7. Total Probability for Sum=4: 3/36 = 1/12.
   Total Probability for Sum=7: 6/36 = 1/6.

8. Final Probability: The game stops with sum=4 given it stops at sum=4 or 7:

   $$\frac{\text{Probability of sum=4}}{\text{Probability of sum=4 or 7}}=\frac{1\mathrm{/}12}{1\mathrm{/}12+1\mathrm{/}6}=\frac{1\mathrm{/}12}{3\mathrm{/}12}=\frac{1}{3}\mathrm{.}$$

9. $$=\frac{1/12}{3/12}$$

10. $$=\frac{1}{3}$$

Thus, the probability is 1/3.