Answer to Question #161992 in Statistics and Probability for uyuh

Question

Suppose you just purchased a digital music player and have put 10 tracks on it. After listening to them you decide that you like 4 of the songs. With the random feature on your​ player, each of the 10 songs is played once in random order. Find the probability that among the first two songs played

​(a) You like both of them. Would this be​ unusual?

​(b) You like neither of them.

​(c) You like exactly one of them.

​(d) Redo​ (a)-(c) if a song can be replayed before all 10 songs are played.

Answer 1

(a) We calculate the number of ways to play the first two songs:

The first song can be chosen by 10 different ways. The chosen song can not be chosen in the second time, therefore the second song can be chosen by 9 different ways. By using multiplication rule, we have that there are $10\cdot 9=90$ ways to play 2 different songs from the list of 10 songs.

Similarly, there are $4\cdot 3=12$ ways to play 2 different songs from the list of 4 my favorite songs. The probability that the both songs played are my favorite is 12/90=0.1333 or 13.33%.

(b) Again, there are $6\cdot 5=30$ ways to play 2 different songs from the list of 6 my non-favorite songs. The probability that the both songs played are my non-favorite is 30/90=0.3333 or 33.33%.

(c) We calculate now the number of ways to play exactly one of my favorite songs among the first two songs played.

A favorite song can be chosen by 4 different ways. A non-favorite song can be chosen by 6 different ways. By using multiplication rule, we have that there are $6\cdot 4=24$ ways to choose such 2 songs. But we can also change the order in which the songs are playing. This doubles number of ways, and the total number becomes 48 ways.

The corresponding probability is 48/90=0.5333 or 53.33%.

Another way to calculate this probability is to use the results of (a) and (b). The events in the cases (a) , (b) and (c) are mutually exclusive and form a complete system of events, therefore, the sum of their probabilities is equal to 1.

The considered probability is 1 - 12/90 - 30/90 = 48/90=0.5333.

(d) The first song can be chosen by 10 different ways. The chosen song can be chosen in the second time, therefore the second song can be chosen by 10 different ways too. By using multiplication rule, we have that there are $10\cdot 10=100$ ways to play 2 different songs from the list of 10 songs.

Similarly, there are $4\cdot 4=16$ ways to play 2 different songs from the list of 4 my favorite songs. The probability that the both songs played are my favorite is 16/100=0.16 or 16%.

Again, there are $6\cdot 6=36$ ways to play 2 different songs from the list of 6 my non-favorite songs. The probability that the both songs played are my non-favorite is 36/100=0.36 or 36%.

The probability that exactly one of my favorite songs will be among the first two songs played is complementary to the first two above: 1-0.16-0.36=0.48 (=48%)

Answer. (a) 13.33%, (b) 33.33%, (c) 53.33%, (d) 16%, 36% and 48%.