AMC 8 1998 Problem 19

AMC8 1998 年第 19 题

概率★★★☆☆

Tamika selects two different numbers at random from the set

\{ 8,9,10 \}

and adds them. Carlos takes two different numbers at random from the set

\{3, 5, 6\}

and multiplies them. What is the probability that Tamika's result is greater than Carlos' result?

Tamika 从集合

\{8,9,10\}

中随机选两个不同的数并相加。Carlos 从集合

\{3,5,6\}

中随机选两个不同的数并相乘。Tamika 的结果大于 Carlos 的结果的概率是多少？

正确答案 · Correct Answer

解析 · Solution

Solution 1

The different possible (and equally likely) values Tamika gets are:

8+9=17, 8+10=18, 9+10=19

The different possible (and equally likely) values Carlos gets are:

3×5=15, 3×6=18, 5×6=30

The probability that if Tamika had the sum 17 her sum would be greater than Carlos's set is

\displaystyle \frac{1}{3}

, because 17 is only greater than 15

The probability that if Tamika had the sum 18 her sum would be greater than Carlos's set is

\displaystyle \frac{1}{3}

, because 18 is only greater than 15

The probability that if Tamika had the sum 19 her sum would be greater than Carlos's set is

\displaystyle \frac{2}{3}

, because 19 is greater than both 15 and 18

Each sum has a

\displaystyle \frac{1}{3}

possibility of being chosen, so we have

\displaystyle \frac{1}{3} \left( \frac{1}{3} + \frac{1}{3} + \frac{2}{3} \right)= \frac{1}{3} \left( \frac{4}{3} \right)= \frac{4}{9} = A

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