Two dice are thrown simultaneously what is the probability that sum of the two numbers will be 5 before 7?
Step-by-Step Solution
Key Concept: This is a geometric series problem where the game repeats with probability 13/18 (neither 5 nor 7 appears). The probability of getting sum 5 before 7 is P(5)/(1 - P(neither)) = (1/9)/(5/18) = 2/5, using the formula for infinite geometric series: a/(1-r) where a = 1/9 and r = 13/18.
The probability of rolling 5 is $4/36 = 1/9$ and rolling 7 is $6/36 = 1/6$. The probability neither 5 nor 7 appears is $1 - 10/36 = 13/18$. The required probability is $\frac{1}{9} + \frac{13}{18} \cdot \frac{1}{9} + \ldots = \frac{1/9}{1-13/18} = \frac{1/9}{5/18} = \frac{2}{5} = 0.40$.
Correct Answer: 0.40