Second Number Larger Than The First

Suppose a six-sided die has sides numbered 1 through 6. If a person throws the die two times, what is the probability that the second number will be larger than the first?
Source: NCTM Mathematics Teacher, September 2006

Solution
The desirable outcomes are
1:\left \{2,3,4,5,6\right \} where 1 is the first number and 2,3,4,5,6 are the second number
2:\left \{3,4,5,6\right \}
3:\left \{4,5,6\right \}
4:\left \{5,6\right \}
5:\left \{6\right \}
6:\left \{\textrm{not possible}\right\}
Total number of desirable outcomes
5+4+3+2+1=15
Number of possible outcomes
6\times 6=36
Probability that the second number is larger than the first = 15/36

Answer: 15/36

Alternative solution
First, the die is thrown and the first number \left \{1,2,3,4,5,6\right \} comes up each with a probability equal 1/6. Then, the die is thrown again and we want the second number larger than the first.
1:\left \{2,3,4,5,6\right \} probability = 5/6
2:\left \{3,4,5,6\right \} probability = 4/6
3:\left \{4,5,6\right \} probability = 3/6
4:\left \{5,6\right \} probability =2/6
5:\left \{6\right \} probability = 1/6
6:\left \{\textrm{not possible}\right \} probability = 0/6
Probability of rolling the die two times and the second number is larger than the first
\dfrac{1}{6}\times \dfrac{5}{6}+\dfrac{1}{6}\times \dfrac{4}{6}+\dfrac{1}{6}\times \dfrac{3}{6}+\dfrac{1}{6}\times \dfrac{2}{6}+\dfrac{1}{6}\times \dfrac{1}{6}+\dfrac{1}{6}\times \dfrac{0}{6}=\dfrac{1}{6}\left (\dfrac{5+4+3+2+1+0}{6}\right )
=\dfrac{1}{6}\times \dfrac{15}{6}
=15/36

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About mvtrinh

Retired high school math teacher.
This entry was posted in Problem solving and tagged , , , , , , , . Bookmark the permalink.

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