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

Answer:

(A) 0.4196

(B) 0.2398

(C) 0.0020

Step-by-step 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 = $^6C_2\times ^9C_3$

Since,

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

Thus, the probability of choosing 3 females and 2 males = $\frac{ ^6C_2\times ^9C_3}{^{15}C_5}$

$=\frac{\frac{6!}{2!4!}\times \frac{9!}{3!6!}}{\frac{15!}{10!5!}}$

= 0.4196

Similarly,

(B)

The probability of choosing 3 liked songs = $\frac{ ^6C_3\times ^9C_2}{^{15}C_5}$

$=\frac{\frac{6!}{3!3!}\times \frac{9!}{2!7!}}{\frac{15!}{10!5!}}$

= 0.2398

(C)

The probability of choosing 5 liked songs = $\frac{ ^6C_5\times ^9C_0}{^{15}C_5}$

$=\frac{\frac{6!}{5!1!}}{\frac{15!}{5!10!}}$

≈ 0.0020