At a certain restaurant, a meal consists of 2 different meats chosen

Question

At a certain restaurant, a meal consists of 2 different meats chosen from pork, fish, chicken and beef, and 1 dessert chosen from cake, pie and ice cream. 
If two brothers randomly choose the items of their meals, what is the probability they order the same meal?

(A) 1/48
(B) 1/36
(C) 1/24
(D) 1/18
(E) 1/12

PyjamaScientist · Answer

We need to find the probability of both having the same meal.
 
In how many ways can the first brother make one meal out of four meats and three dessert options?
It is  4c2*3c1 = 18.

So, the probability that he chooses either one of the  18  possibilities =  \frac{1}{18} .

Now the second brother has only "one" choice which is the meal that the first brother has chosen. So, the probability for him is  \frac{1}{1} , because he has no other option for both have to have the same meal.

Thus, total probability =  \frac{1}{18}*1 = \frac{1}{18}.

ashishpathak · Answer

The second brother can order any meal that is my understanding isnit correct ? Why should he have only one choice ?

Posted from my mobile device

PyjamaScientist · Answer

ashishpathak ,

Because both brothers have to have the "same meal". So after the first brother has chosen one combination of the meal, second brother's meal "has to be that same meal". So, for him favourable case= 1 as well as total cases also = 1. Probability = 1/1.

Hope it helps.

belagavi · Answer

There are 6 (4c2) meat combinations and 3 ice cream options. Therefore there could be a total of 18 combinations of meat and ice cream together.
Let the first brother choose any one of the 18 combinations. 
Now, the second brother also has to choose his food. He has 18 combinations to choose from. 
But what is the probability that he chooses the same combination as his brother?
Its 1 out of 18.

ThatDudeKnows · Answer

ashishpathak

I'm guessing you were thinking about this from the other direction, which is completely fine.

There are 4!/2! ways to pick the meats and 3 ways to pick the dessert, so (4*3*3)/2 = 18 ways to pick the meal.

The first brother picks a meal.

Now, the second brother must match what the first brother picked. There is one way for him to do that out of 18 possible meals he could have assembled.

So, 1/18.

Answer choice D.

Note that this is the same thing that  PyjamaScientist  did, it's just that he described it as the first brother matching the second whereas I described it as the second brother matching the first. Same thing!

ashishpathak · Answer

thanks clear now nic eexplanation

BrentGMATPrepNow · Answer

Let's first calculate the total number of possible meals one can order at the restaurant.

The number of ways to choose 2 different meats from 4 options = 4C2 = (4)(3)/(2)(1) =   6   
The number of ways to choose 1 dessert from 3 options =   3   
So, the TOTAL number of possible meals the first brother can choose from = (  6  )(  3  ) =   18

Once the first brother has chosen his meal, there's only 1 way (out of   18   options) the second brother's selection can match the first brother's selection.

So, P(both brothers order the same meal) = 1/  18

Answer: D

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At a certain restaurant, a meal consists of 2 different meats chosen

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At a certain restaurant, a meal consists of 2 different meats chosen

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