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1st plate can be any plate. It is the second plate that has to be picked such that it matches..

Thus, 15/15 * 4/14 (4 remaining out of the set whose first plate got chosen, and 14 total remaining as 1 plate is already chosen)

2/7
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Given
    • A box at a yard sale contains 3 different china dinner sets.
      o Each set consists 5 plates.
    • A customer randomly selects 2 plates to check for defects

To find

    • The probability that the 2 plates selected will be from the same dinner set.

Approach
    • Probability that the 2 plates selected will be from the same dinner set = Total ways of picking 2 plates from the same set / Total ways of picking 2 plates
      o Total ways of picking 2 plates from the same set = 3 × 5 C 2 = 30
      o Total ways of picking 2 plates = 15 C 2 = 105
    • Probability that the 2 plates selected will be from the same dinner set = \(\frac{30 }{105}\) = \(\frac{2 }{7}\)
Hence, the correct answer is option A.
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Given: A box at a yard sale contains 3 different china dinner sets, each consisting of 5 plates. A customer will randomly select 2 plates to check for defects.
Asked: What is the probability that the 2 plates selected will be from the same dinner set?

Total number of ways to select 2 plates = 15C2 = 105
Favorable ways = 3*5C2 = 3*10 = 30

The probability that the 2 plates selected will be from the same dinner set = 30/105 = 2/7

IMO A
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Hello from the GMAT Club BumpBot!

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