Question

I have 100 albums on my cellphone, each album has 10 songs in it. My phone randomly picks a song and plays it. After playing it, it repeats the process. If each song is 5 minutes long, then what's the probability:

a. Plays 2 or more songs from the same album in the next 1 hour?

b. Plays 2 or more songs from the same album in the next 2 hours?

Answer 1

It is with replacement because the phone doesn't exclude already played song before playing the next song.

Thus, the binomial distribution has to be used, here.

P(X =r) =C(n, r).p^r.q^n-r

(where, P =Probability; X =r =number of successes (X is a binomial random variable); p =proportion of success on single trial; q =proportion of failure on single trial =1 - p; n =Sample size).

a)

1 hour =60 minutes =12*5 minutes =12 songs.

So, the sample size is n =12

X =2 or more songs

p =proportion of 'success', i.e., plays a song from a 'particular album' =1 song/10 songs =1/10 =0.1

(because each album has 10 songs in it and so, a song from any particular album means it is out of 10 songs in that album).

P(X =2 or more songs from the same album) =P(X 2) =1 - P(X < 2) =1 - [P(X =0)+P(X =1)] =0.3410

b)

Everything is same as in part (a) above except for n =24 songs because 2 hours =2*60 mins =2*12*5 mins =24*5 mins =24 songs.

So, n =24; p =0.1; X =2 or more

P(X =2 or more songs from the same album) =0.7075