The names of 3 songs are placed in each of 4 bags and one

Question

The names of 3 songs are placed in each of 4 bags and one name is drawn out of each bag. What is the probability that exactly 2 songs will be included in the resulting 4-song sequence?

(A) 7/18 (B) 4/9 (C) 1/2 (D) 14/27 (E) 5/9

durgesh79 · Answer

B - 4/9

Suppose three songs are A,B,C 
4 bags, each can have 3 songs selected.

total possibilities = 3^4

no of ways a perticular song can be picked = 3
no of ways it can be in exactly 2 bags = 6 (1,2)(1,3)(1,4....... ) 4C2
other two songs can take 2 places in rest of the two bags in 2 ways

total fav = 3*6*2

probablility = fav/total = 3*6*2/3*3*3*3 = 4/9

Oski · Answer

First of the two song : 3 possibilities
Second of the two songs : 2 possibilities
=> 6 (A,B) possibilities (ordered pair)

By symmetry, we can fix the first song to be A (to avoid double counting):

Three possibilities for two of each:
AABB
ABAB
ABBA

Three possibilities for three A and one B:
AAAB
AABA
ABAA

Total 6*(3+3) = 36 possibilities out of the 3^4=81 that exist

==> Answer is 36/81 = 4/9 i.e. (B)

Edit:

there is also the possibility for one A and three B:
ABBB

So total is 6*(3+3+1) = 42 possibilities out of the 3^4=81 that exist

==> Answer is 42/81 = 14/27 i.e. (D)

maratikus · Answer

the answer is D

total number of scenarios: 3^4 = 81
good scenarios: kick one song out (3 ways), there are 16 scenarios left (2^4) two of which are bad aaaa and bbbb -> 3*14/81 = 14/27

Oski · Answer

I think you have the right answer but I don't get the end of the reasonning.

Why would there be 16 scenarii left ?

If you fixed the first song A (seems to be what you did by saying "kick one song out (3 ways)"), you just have 3 songs left to choose. And where did you choose the second song in your method? It seems you immediately assume it is song B, but it could be the third one.

Applying your method, I would say:

Kick the first song out -> 3 ways to choose A the first song
Choose the second song -> 2 ways to choose B the other song
There are 8 scenarios left for the 3 last songs (2^3) one of which is bad (AAA) -> 7 ways

Total = 3*2*7 / 81 = 14/27

In my solution above, I forgot the "One A and three B" possibility, which would have led me to 6*(3+3+1)/81 = 14/27 too

durgesh79 · Answer

thanks maratikus. good approach... i totally misunderstood the Question.

apurva1985 · Answer

Dont we need consider taking the third Song out of the bag as we have 3 songs to select from :?:

durgesh79 · Answer

as per question stem, we need to take events, when only (and exactly) two songs are picked...

as explained above, there are 14 ways you can pick two songs. Now these songs can be AB, BC or CA (3 ways)

total fav = 3*14

The names of 3 songs are placed in each of 4 bags and one

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