Question

a disc jockey has 8 songs:2 rock songs3 reggae3 country songsthe disc jockey Randomly chooses the 1st song to play, then Randomly chooses the 2nd song from the remaining ones.what is the Probability that 1st song is a Rock song and the 2nd is a country song?write a fraction- simplest form

Answer 1

Solution:

Given:

$\begin{gathered} A\text{ disc jockey has 8songs} \\ \\ \text{Rock songs = 2} \\ \text{Reggae = 3} \\ \text{Country = 3} \\ \\ \text{TOTAL = 8} \end{gathered}$

Since the disc jockey randomly chooses the first song to play and then randomly chooses the second song from the remaining ones, this is a probability without replacement.

Hence, the probability that the first song is a rock song and the second is a country song will be;

$\begin{gathered} \text{Let R = rock song} \\ \text{Let C = country song} \\ \\ \text{Thus,} \\ P(RC)=P(R)\times P(C) \\ P(\text{Rock)}=\frac{n\text{ umber of rock songs}}{\text{total songs}} \\ P(R)=\frac{2}{8}=\frac{1}{4} \\ \text{Total songs left after the first song has b}een\text{ played is 7} \\ P(country\text{)}=\frac{n\text{ umber of country songs}}{\text{total songs left}} \\ P(C)=\frac{3}{7} \end{gathered}$

Thus,

$\begin{gathered} P(RC)=P(R)\times P(C) \\ P(RC)=\frac{1}{4}\times\frac{3}{7} \\ P(RC)=\frac{3}{28}_{} \end{gathered}$

Therefore, the probability that the first song is a rock song and the second is a country song as a fraction in its simplest form is;

$\frac{3}{28}$