2015 AMC 8

Complete problem set with solutions and individual problem pages

Problem 7 Easy

Each of two boxes contains three chips numbered 1, 2, 3. A chip is drawn randomly from each box and the numbers on the two chips are multiplied. What is the probability that their product is even?

  • A.

    \frac19

  • B.

    \frac29

  • C.

    \frac49

  • D.

    \frac12

  • E.

    \frac59

Answer:E

Solution 1

You can make this problem into a spinner problem. You have the first spinner with 3 equally divided

sections: 1, 2, and 3. You make a second spinner that is identical to the first, with 3 equal sections of

1,2, and 3. If the first spinner lands on 1, it must land on two for the result to be even. You write down the first

combination of numbers: (1,2). Next, if the spinner lands on 2, it can land on any number on the second

spinner. We now have the combinations of (1,2) ,(2,1), (2,2), and (2,3). Finally, if the first spinner ends on 3, we

have (3,2). Since there are 3\cdot3=9 possible combinations, and we have 5 evens, the final answer is

\boxed{\textbf{(E) }\frac{5}{9}}.

 

Solution 2

We can list out the numbers. Box A has chips 1, 2, and 3, and Box B also has chips 1, 2, and 3. Chip 1 (from Box A)

could be with 3 partners from Box B. This is also the same for chips 2 and 3 from Box A. 3+3+3=9 total sums. Chip 1 could be multiplied with 2 other chips to make an even product, just like chip 3. Chip 2 can only multiply with 1 chip. 2+2+1=5. The answer is \boxed{\textbf{(E) }\frac{5}{9}}.

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