Taro rolls a die and then Jiro rolls a die next.
In this case, find the probability that Taro's number is larger than Jiro's number.
Answer
The probability that the number of Taro is larger than that of Jiro is the same as the probability that the number of Jiro is larger than that of Taro.
Then the probability that the number of Taro is larger than that of Jiro is 5/12 which is half of 5/6.
The case the number of Taro is larger than that of Jiro is as follows :
In total 5 + 4 + 3 + 2 + 1 = 15 ways.
Then the probability is calculated as 15/36=5/12.
5/12
Solution
The probability when I roll two dice and the number of one is larger than another is 5/6, which is the probability 1/6 of the same number of both dice is subtracted from 1.
The probability that the number of Taro is larger than that of Jiro is the same as the probability that the number of Jiro is larger than that of Taro.
Then the probability that the number of Taro is larger than that of Jiro is 5/12 which is half of 5/6.
<Another solution>
All number of combination is 6 × 6 = 36.
The case the number of Taro is larger than that of Jiro is as follows :
Taro's is 6 and Jiro's is 5~1,
Taro's is 5 and Jiro's is 4~1,
Taro's is 4 and Jiro's is 3~1,
Taro's is 3 and Jiro's is 2~1,
Taro's is 2 and Jiro's is 1.
In total 5 + 4 + 3 + 2 + 1 = 15 ways.
Then the probability is calculated as 15/36=5/12.
