A box at a yard sale contains 3 different china dinner sets, each cons

Question

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. What is the probability that the 2 plates selected will be from the same dinner set?

A. \(\frac{2}{7}\)

B. \(\frac{2}{5}\)

C. \(\frac{2}{3}\)

D. \(\frac{5}{6}\)

E. \(\frac{3}{2}\)

A.  \frac{2}{7}

B.  \frac{2}{5}

C.  \frac{2}{3}

D.  \frac{5}{6}

E.  \frac{3}{2}

gmatlover12 · Answer

In how many ways can s/he select 2 plates? There are 15 plates, so  15C_2  = 105

In how many ways can s/he select 2 plates from the same set? This = (number of ways of selecting the set) x (number of ways of selecting 2 plates within that set)

= (3) x ( 5C_2 )
= 3 x 10 = 30

Thus, the probability is  \frac{30}{105}  =  \frac{2}{7}

ScottTargetTestPrep · Answer

Since there are 3 different dinner sets and each set has 5 plates, there are a total of 15 plates. So the total number of ways to choose 2 plates is 15C2 = 15!/(2! x 13!) = (15 x 14)/2! = 15 x 7 = 105.

We can let the 3 dinner sets be set 1, set 2, and set 3. The number of ways to choose 2 plates from set 1 is 5C2 =5!/(2! x 3!) = (5 x 4)/2! = 5 x 2 = 10.

Since there are also 10 ways to choose 2 plates from each of the other two sets, there are a total of 30 ways to choose 2 plates from the same dinner set.

Thus the probability is 30/105 = 2/7.

Alternate Solution:

Since there are 3 different dinner sets and each set has 5 plates, there are a total of 15 plates. After the first plate is selected, there are 4 other plates belonging to the same set as the selected plate and 14 remaining plates. Thus, the probability that the second selected plate is the same kind as the first selected plate is 4/14 = 2/7.

Answer: A

wishmasterdj · Answer

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

EgmatQuantExpert · Answer

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.

Kinshook · Answer

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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A box at a yard sale contains 3 different china dinner sets, each cons

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A box at a yard sale contains 3 different china dinner sets, each cons

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