Question

Suppose a compact disk​ (CD) you just purchased has 1515 tracks. After listening to the​ CD, you decide that you like 66 of the songs. The random feature on your CD player will play each of the 1515 songs once in a random order. Find the probability that among the first 55 songs played​ (a) you like 2 of​ them; (b) you like 3 of​ them; (c) you like all 55 of them.

Answer 1

(A) 0.4196

(B) 0.2398

(C) 0.0020

Explanation:

Given,

Total songs = 15,

Liked songs = 6,

So, not liked songs = 15 - 6 = 9

If any 5 songs are played,

Then the total number of ways =
`^(15)C_5`

(A) Number of ways of choosing 2 liked songs =
`^6C_2* ^9C_3`

Since,

`\text{Probability}=\frac{\text{Favourable outcomes}}{\text{Total outcomes}}`

Thus, the probability of choosing 3 females and 2 males =
`( ^6C_2* ^9C_3)/(^(15)C_5)`

`=((6!)/(2!4!)* (9!)/(3!6!))/((15!)/(10!5!))`

= 0.4196

Similarly,

(B)

The probability of choosing 3 liked songs =
`( ^6C_3* ^9C_2)/(^(15)C_5)`

`=((6!)/(3!3!)* (9!)/(2!7!))/((15!)/(10!5!))`

= 0.2398

(C)

The probability of choosing 5 liked songs =
`( ^6C_5* ^9C_0)/(^(15)C_5)`

`=((6!)/(5!1!))/((15!)/(5!10!))`

≈ 0.0020