Maria had five boxes, same in design and size but different in color.

Question

Maria had five boxes, same in design and size but different in color. She removed the lids of all boxes and randomly put them back. What is the probability that exactly two boxes had the matching lids?

A. 1/12
B. 1/6
C. 1/4
D. 1/3
E. 1/2

A. 1/12
B. 1/6
C. 1/4
D. 1/3
E. 1/2

Bunuel · Accepted Answer

We can choose the two boxes out of five that will have the matching lids in 5C2 ways. We need the remaining three boxes to have not matching lids; this can be done only in the following two ways (much better to simply list the possibilities than trying to come up with a formula because the numbers are small): A - B - C (boxes) b - c - a (lids) c - a - b (lids) Thus, the total number of ways to match the lids to only two boxes in 5C2*2=20. The total number of ways to assign the lids to the boxes is simply 5!. Therefore, the probability is 20/120 = 1/6. Answer: B. Similar questions to practice: https://gmatclub.com/forum/tanya-prepar ... 88626.html https://gmatclub.com/forum/letter-arran ... 84912.html https://gmatclub.com/forum/micky-has-10 ... 13801.html https://gmatclub.com/forum/5-letters-ha ... 89501.html https://gmatclub.com/forum/five-letters ... 24028.html Hope it helps.

Rexus · Answer

1) sample space = 5! (ways of placing lid on boxes) 
2) Favourable outcome Maria places 2 matching lid 
 that would be 5c2. 
now we also have reamaining 3 boxes, they will be placed in 3! ways but 
 there is a chance of all three matching the box so we will subtract 1
there is also a chance of any one matching lid so we will subtract 1*3. {(3 lids )}

favourable outcome is 5c2*(3!-3-1) =20

answer is 20/120 =1/6

eleihu · Answer

total outcomes 5!=120. 
first two matched, 
C(5,2)= 
=P(5 , 2)/ P(2 , 2)
=5!/ (5 - 2)!/ 2!
=5! /3! /2!
=20/2
=10

remaining three dont match, first has 2 outcomes, second has 1, third has 1, than 10x2X1x=20

20/120= 1/6.

the winner goes to B. am i right?

Bunuel · Answer

EXPERT'S GLOBAL OFFICIAL VIDEO EXPLANATION

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MT1302 · Answer

Hey  Bunuel

Can you explain your way of solving this question ?

siddhantvarma · Answer

Understanding the problem took some time. We have 5 boxes, different color so basically all boxes are different, none are identical.

Let the boxes be: B1, B2, B3, B4, B5
And their respective lids be: L1, L2, L3, L4, L5

The lids of all boxes are removed. We need to find the probability that exactly two boxes had matching lids.

The problem can be seen as, In how many ways can you assign ONLY 2 boxes their respective lids such that those two boxes get their respective lids, but all the others DO NOT
For ex, one possibility is B1 gets L1 and B2 gets L2 (both B1 and B2 get their respective lid), but B3 gets L4, B4 gets L5 and B5 gets L3
 
You’re arranging the lids, from L1 to L5 on top of boxes, from B1 to B5
Total number of ways to arrange the lids: 5! (5 boxes, 5 lids) = 120

Number of ways to correctly assign only two lids and incorrectly assign the remaining three lids: 
- First, select those two lids in 5C2 = 10 ways 
- Now, these two lids can only be arranged in 1 way, since each lid needs to be placed on top of it’s respective box 
- We now need to arrange or basically “de-arrange” the remaining 3 lids on 3 boxes such that none of the lids are placed on top of their respective boxes
- The only possible scenarios are:
 - L4, L5, L3
 - L5, L3, L4
- Hence there are two ways to de-arrange the remaining three lids 
- Total number of ways to assign two lids correctly and the remaining incorrectly: 10x2 = 20

Probability: 20/120 = 1/6 (B)
 
Here's a little 2 min concept notes on "de-arrangement":

- A derangement is a permutation where no element stays in its original position.
- For example, if we have items A, B, C:
 - One possible permutation: C, A, B
 - If none of A, B, or C are in their original positions, it's a derangement.

There's also a complicated recursive formula for arrangements, but I don't think we'll need that in GMAT. Instead the given table should be more than enough. I'd memorise the values till n=4 or n=5 but not beyond that:

n Dₙ
1 0
2 1
3 2
4 9
5 44
6 265
7 1854
8 14833
9 133496
10 1334961

The above table shows the number of ways to de-arrange n different objects given by Dₙ

GMATinsight · Answer

Total ways of putting lids back on 5 boxes randomly = 5*4*3*2*1 = 5! = 120

Favorable outcomes = 2 boxes correct lid and other three wrong ones

2 box with correct lid may be kept separate in 5C2 = 10 ways

3 lids on remaining three boxes can be put incorrectly in following way
first lid can be put on any two of the wrong boxes
second lid can be put on only one box incorrectly so that third also remains incorrect
so outcomes = 2*1*1

Total favorable outcomes = 5C2*2*1*1 = 20

Probability = 20/120 = 1/6

Answer: Option B

Related Video: 
  https://www.youtube.com/watch?v=G7a_MBqZjTY

Sushi_545 · Answer

Hi  Bunuel ,

can you pls comment whats wrong here:-

1/5* 1/4 * 2/3 *1/2 *1

prob (1st box with correct lid )= 1/5
 prob (2nd box with correct lid )= 1/4
 prob (3rd box with incorrect lid )= 2/3
 prob (4th box with in correct lid )= 1/2
 prob (5th box with incorrect lid )= 1/1

Regor60 · Answer

This approach assumes a specific set of two boxes have the correct lids.

The question is asking for ANY set of two boxes.

Since there are 5!/3!2!=10 ways to select two boxes, the probability is 10 times your approach (1/60), or 1/6

bhanu29 · Answer

This should work, if u consider choosing which 2 boxes have lids correctly placed

So it will be 5C2 *  1/5* 1/4 * 2/3 *1/2 *1 
10 * 0.2 * 1/4 * 2/3 * 1/2 = 1/6

which is the answer B

minusaccusamus · Answer

Not sure how but I ended up finding the right answer with a formula I am not sure where I learned :

combination of right lids arrangement / permutation of wrong lids arrangement

2C5 / 3P5 = 10 / 60 = 1/6

Is this just a weird coincidence ? I could swear i saw this method somewhere.

Maria had five boxes, same in design and size but different in color.

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Maria had five boxes, same in design and size but different in color.

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