Song Sequence

[kevinspain]

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

[Praso]

D : 14/27
Total number of combinations : 3^4 = 81----- EQ1
no. of ways to choose 2 out of 3 songs = 3c2 = 3 (say songs a and b)--- EQ2

number of ways these 2 songs can be arranged in 4 slots in the sequence, so that a song is present atleast in one slot.
= 2^4 - 2 (2 is for aaaa and bbbb)
=14 ---EQ3

EQ 2 & 3 => 3*14 = 42 number of ways.

So probablity = 42/81 = 14/27!


PS: took me atleast 10 mins to solve this

[krovvidy]

Consider that a song 'a' is selected from bag-1. From then on, the probability of getting the same song 'a' is 1/3 and that of anyother song is 2/3.

So, the probability of there being exactly 2 songs =

1*(1/3)*(1/3)*(2/3) + 1*(1/3)*(2/3)*(2/3) + 1*(2/3)*(1/3)*(2/3) + 1*(2/3)
*(2/3)*(1/3)

= 14/27 (D)

[kevinspain]

Excellent!

[raunekk]

not getting it..

krovvidy please explain once more..

[krovvidy]

Rephrasing the qn:

There are 3 songs a, b & c in each of the 4 bags. One song each would be drawn from each bag at random. Out of the 4 songs drawn, what is the probability that there are exactly 2 distinct songs.

On the first draw (i.e., from bag-1), any of the 3 songs can be drawn (probability = 1).

On the second draw (i.e., from bag-2), we can either draw the same song drawn from bag-1 (probability = 1/3) or any other song (probability = 2/3)

On the third draw (i.e., from bag-3), we can either draw the same song drawn from bag-1 (probability = 1/3) or any other song (probability = 2/3)

On the fouth draw (i.e., from bag-4), we can either draw the same song drawn from bag-1 (probability = 1/3) or any other song (probability = 2/3)

Now you can see how the probability of getting 2 distinct songs out of the 4 draws is given by:

1*(1/3)*(1/3)*(2/3) + 1*(1/3)*(2/3)*(2/3) + 1*(2/3)*(1/3)*(2/3) + 1*(2/3)*(2/3)*(1/3)

HTH

Originally Posted by raunekk

not getting it..

krovvidy please explain once more..

[ASKajan]

Lets say each of four bags contains (Stairway, Wonderwall and Yellow)...

The question is asking....what is the probability that we choose exactly 2 songs on four tries

First part of the equation = 1*(1/3)*(1/3)*(2/3)

1st bag...any song we choose can be one of the two songs so = 1 (lets say wonderwall)
2nd bag...we have a 1/3 chance of choosing wonderwall so = 1/3
3rd bag...again we have a 1/3 chance of choosing wonderwall so = 1/3
4th bag....question asks for 2 songs...we have 3 wonderwalls and need a stairway or yellow so = 2/3

The rest of the equation accounts for other variations, such as:

Wonderwall, Wonderwall, Yellow, Yellow
Wonderwall, Yellow, Wonderwall, Yellow
Wonderwall, Yellow, Yellow, Wonderwall

I used Wonderwall and Yellow as my two choices but the variables could have been called Song A and Song B as this equation accounts for the probability of choosing exactly two distinct songs from 1 pick from each of 4 bags.

[raunekk]

thanks krovvidy.. tat was a gr8 explanation..

[raunekk]

ASKajan vbmenu_register("postmenu_611228", true); ,as u said..

1st bag...any song we choose can be one of the two songs so = 1 (lets say wonderwall..

IS it 1, because there are 3 ways a song can b selected and there are total of 3 songs..therefore 3/3 =1??

Also u said,
4th bag....question asks for 2 songs...we have 3 wonderwalls and need a stairway or yellow so = 2/3

I didnt understood this..

thanks ..

[ASKajan]

3/3 = 1....because our first song has a 100% chance of being 1 of the 2 distinct songs...regardless of the one we choose.

4th bag....I was trying to explain that if we choose wonderwall, wonderwall, and wonderwall from the first 3 bags (1*1/3*1/3) then the only way we can choose two distinct songs (as the question stem tells us to do) is to choose any song from the 4th bag that ISNT wonderwall...better?