At a certain deli, 3/5 of the sandwiches on the menu contain meat as a

Question

At a certain deli, 3/5 of the sandwiches on the menu contain meat as an ingredient and 1/2 of the sandwiches contain cheese. If 7/10 of the sandwiches that contain cheese as an ingredient also contain meat, and a customer orders 2 sandwiches at random from the menu, what is the probability that both sandwiches contain meat, but neither contain cheese?

A. 1/100
B. 7/100
C. 1/16
D. 3/10
E. 1/2

Show Answer

Official Answer

C

A. 1/100
B. 7/100
C. 1/16
D. 3/10
E. 1/2

leeemowlimeow · Answer

Contain cheese and meat = 1/2*7/10=7/20
contain meat but not cheese=contain meat -contain both=3/5-7/20=5/20
Order two containing meat but not cheese=(5/20)*(5/20)=1/16

GMATWhizTeam · Answer

Let the total number of sandwiches on the menu be 100. (we are taking 100 for the ease of work)
Number of sandwiches containing meat =  \frac{3}{5}  of 100 = 60.
Number of sandwiches containing cheese =  \frac{1}{5}  of 100 = 50.

Number of sandwiches containing meat and cheese both =  \frac{7}{10}  of 50 = 35.

Number of sandwiches containing only meat = no. of sandwiches containing meat - no. of sandwiches containing both = 60-35 = 25.

Probability of choosing 2 sandwich containing only meat =  (\frac{25}{100})^2  =  \frac{1}{16}

Hence the correct answer is  \frac{1}{16}

KetakiUpadhye · Answer

Hi GMATWHIZ team. Can u please tell me why u considered 50 sandwiches instead of the previous Quantity of 100 sandwiches.

Posted from my mobile device

vmilare · Answer

Where in this problem is the quote that we have the replacement of the sandwiches? I think we have this gap on the question don't we?

Armaan209 · Answer

Yes it should specify that. In a real scenario it would be awithout replacement. You must have landed to 1/20 answer.

rak08 · Answer

Bunuel

where did I go wrong??

assuming 20

so 12 are meat
 10 are cheese
 7 are both

so only meat = 5

5c2/20c2 = 1/19

Krunaal · Answer

Since you can select the same sandwich from the menu twice, you would do  5/20 * 5/20 = 1/16

When you do 5C2, you select 2 different sandwiches that have meat without cheese; but that's not a condition in the question.

luisdicampo · Answer

Deconstructing the Question

Topic: Conditional Probability & Sets

1. Define Events & Probabilities  
 
  P(M)  = Probability of Meat =  3/5  
  P(C)  = Probability of Cheese =  1/2  
  P(M | C)  = Probability of Meat  given  Cheese =  7/10

2. Find the Intersection: P(M and C)  
Using the conditional probability formula:  P(M \cap C) = P(C) * P(M | C) 
 P(M \cap C) = (1/2) * (7/10) =    7/20

3. Find "Meat Only"  
We want sandwiches with Meat but  excluding  the intersection (Meat & Cheese).
 P(Meat Only) = P(M) - P(M \cap C) 
 P(Meat Only) = 3/5 - 7/20 
 P(Meat Only) = 12/20 - 7/20 =    5/20 = 1/4

4. Final Calculation  
Picking 2 sandwiches independently from the "Meat Only" group:
 P(Total) = (1/4) * (1/4) =    1/16

The correct answer is C.

At a certain deli, 3/5 of the sandwiches on the menu contain meat as a

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At a certain deli, 3/5 of the sandwiches on the menu contain meat as a

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