Question

A radio disc jockey has 8 songs on this upcoming hour's playlist: 2 are rock songs, 4 are reggae songs, and 2 are country songs. The disc jockey randomly chooses the first song to play, and then she randomly chooses the second song from the remaining ones. What is the probability that both songs are reggae songs? Write your answer as a fraction in simplest form.

Answer 1

Answer:

$\displaystyle \frac{1}{28}$

Step-by-step explanation:

we are given 8 songs

2 are rock. 4 are reggae, 2 are country

we want to figure out the probability of getting both reggae songs

notice that, they aren't replacing thus it's an independent probability

recall that,

$\displaystyle \rm P(A \: \text{and} \: B) = P(A) \times P(B \mid A)$

P(A) represents the chance of getting our subject first time and P(B|A) represents the chance of getting the same subject second time

let's figure out P(A)

we have 2 reggae songs

and total 8 songs

therefore

$\displaystyle\rm P(A) = \frac{2}{8} = \frac{1}{4}$

let's figure out P(B|A)

since we took a reggae song before now we have only a reggae song

and total 8-1=7

therefore

$\displaystyle \rm P(B\mid A) = \frac{1}{7}$

Altogether

$\displaystyle \rm P(A \: \text{and} \: B) = \frac{1}{4} \times \frac{1}{7}$

simplify multiplication:

$\displaystyle \rm P(A \: \text{and} \: B) = \frac{1}{28}$

And we are done!