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25 May 2018, 04:11
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A
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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
Joined: 25 Jan 2017
Last visit: 11 Dec 2021
Posts: 26
Given Kudos: 9
Location: India
Concentration: Finance, Strategy
Schools: Tuck '21 (A) CBS '21 (II) Kellogg '21 (M$)
GMAT 1: 750 Q50 V41
GMAT 2: 770 Q51 V45
WE:Analyst (Finance: Venture Capital)
25 May 2018, 05:20
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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 \(\frac{30}{105}\) = \(\frac{2}{7}\)
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 \(\frac{30}{105}\) = \(\frac{2}{7}\)
29 May 2018, 09:20
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Bunuel
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. \(\frac{3}{2}\)
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
