Question

a radio disc jockey has 8 songs. on the upcoming playlist:2 rock songs3 reggae songs3 country songsthe disc jockey Randomly choose the 1st song to play, then Randomly choose the 2nd song from the remaining ones. what is the Probability that the 1st song is a Rock song? the 2nd song is a Country song?* write as fraction

Answer 1

Probabilities

We have the following elements to choose from:

Rock songs (R) = 2

Reggaes songs (G) = 3

Country songs (C) = 3

It's required to find two probabilities:

a) The first is a rock song

As stated above, there are R = 2 rocks songs and there are 8 songs in total, thus the probability of randomly picking a rock song is:

`\begin{gathered} p(R)=(2)/(8) \\ \text{Simplify:} \\ p(R)=(1)/(4) \end{gathered}`

b) The second song is a country song.

This is quite more complex to calculate. Let's assume the first song was not a Country song (NC). This has a probability of:

`P(NC)=(5)/(8)`

Since there are 5 non-country songs out of 8. For the second pick, we have all of the country songs unplayed out of 7, thus the probability of selecting it is:

`P(C)=(3)/(7)`

The probability of this 'brank' of events is:

`P(NC,C)=(5)/(8)\cdot(3)/(7)=(15)/(56)`

Now assume the first song was actually a country song. It has a probability of:

`P(C)=(3)/(8)`

For the second pick, we now have only 2 country songs out of 7, so the probability of selecting another country song is:

`P(C)=(2)/(7)`

The probability of this 'branch' of events is:

`\begin{gathered} P(C,C)=(3)/(8)\cdot(2)/(7)=(6)/(56) \\ Simplify\colon \\ P(C,C)=(3)/(28) \end{gathered}`

The total probability of selecting a country song as the second song is:

`\begin{gathered} P(B)=(15)/(56)+(3)/(28) \\ P(B)=(15)/(56)+(6)/(56) \\ P(B)=(21)/(56) \\ P(B)=(3)/(8) \end{gathered}`