Answer to Question #104516 in Statistics and Probability for Joseph

Question

1. You have downloaded 20 new songs onto your phone as shown in the table below
A Tribe Called Red : 3
Drake : 8
The weekend : 7
Alessia Cara : 2

Assuming order does not matter , if you randomly play 8 from your new list of downloaded songs, what is probability, without evaluating, that you will hear
a) no A tribe called red songs?
b) exactly three Drake songs or exactly four Drake songs
c) two songs from each artist
d) at least one The weekend songs

Answer 1

$\begin{matrix} Artist & Number\ of\ songs \\ A\ Tribe \ Called \ Red & 3 \\ Drake & 8\\ The \ weekend & 7 \\ Alessia \ Cara& 2 \end{matrix}$

The possible number of ways for playing 8 songs out of 20 from the list is:

$$\binom{20}{8}={20! \over 8!(20-8)!}={20\cdot19\cdot18\cdot17\cdot16\cdot15\cdot14\cdot13 \over 1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8}=125970$$

a) no A tribe called red songs? 

$$P(no\ A\ Tribe \ Called \ Red )={\binom{20-3}{8} \over\binom{20}{8}}={{17! \over 8!(17-8)!}\over125970}=$$ $$={24310 \over 125970}={11 \over 57}\approx0.19298$$

b) exactly three Drake songs or exactly four Drake songs  

$$P(exactly \ three\ Drake\ songs )={\binom{8}{3}\binom{20-8}{8-3} \over\binom{20}{8}}=$$ $$={{8! \over 3!(8-3)!}\cdot{12! \over 5!(12-5)!}\over125970}={56\cdot792 \over 125970}\approx0.35208$$

$$P(exactly \ four\ Drake\ songs )={\binom{8}{4}\binom{20-8}{8-4} \over\binom{20}{8}}=$$ $$={{8! \over 4!(8-4)!}\cdot{12! \over 4!(12-4)!}\over125970}={70\cdot495 \over 125970}\approx0.27507$$

$$P(exactly \ three\ Drake\ songs\ or\ exactly \ four\ Drake\ songs)=$$ $$={44352 \over 125970}+{34650 \over 125970}={79002 \over 125970}\approx0.62715$$

c) two songs from each artist  

$$P(two\ songs\ from\ each\ artist )={\binom{3}{2}\binom{8}{2}\binom{7}{2}\binom{2}{2} \over\binom{20}{8}}=$$ $$={{3! \over 2!(3-2)!}\cdot{8! \over 2!(8-2)!}\cdot{7! \over 2!(7-2)!}\cdot{2! \over 2!(2-2)!}\over125970}={3\cdot28\cdot21\cdot1\over 125970}\approx0.01400$$

d) at least one The weekend songs

$$P(at\ least\ one \ The\ weekend\ songs)=1-P(no\ The\ weekend\ songs) =$$ $$=1-{\binom{20-7}{8} \over\binom{20}{8}}=1-{{13! \over 8!(13-8)!}\over125970}=1-{1287 \over 125970}\approx0.98978$$