Data Sufficiency

A minidisk player is loaded with only rock, jazz, and opera songs. If the probability of 3 rock songs playing in a row is the same as the probability of 2 jazz songs playing in a row, what is the probability that the first 2 songs played will be an opera song and a rock song?
A 1/64
5/32

B

C 1/32




2

varun289 wrote:A minidisk player is loaded with only rock, jazz, and opera songs. If the probability of 3 rock songs playing in a row is the same as the probability of 2 jazz songs playing in a row, what is the probability that the first song played will be an opera song and the second song played will be a rock song?

1/64

1/32

5/64

1/9

5/32

I've amended the question stem to clarify what I believe is its intent.

Let R = rock, J = jazz, O = opera, and T = the total number of songs.
P(R) = R/T.
P(J) = J/T.

Since the probability that 3 rock songs play in a row is the same as the probability that 2 jazz songs play in a row, we get:
P(RRR) = P(JJ).
(R/T)(R/T)(R/T) = (J/T)(J/T)
(R/T)³ = (J/T)².

Plug in values that satisfy the equation above:
(1/4)³ = (1/8)².
Thus, if 1/4 of the songs are Rock, and 1/8 of the songs are Jazz, the constraint that P(RRR) = P(JJ) is satisfied.

Let the total number of songs = 8.
Number of rock songs = (1/4)8 = 2.
Number of jazz songs = (1/8)8 = 1.
Number of opera songs = 8-2-1 = 5.

The question stem asks for the probability that the first song played will be opera and the second song played will be rock:
P(OR) = (O/T)(R/T) = (5/8)(2/8) = 5/32.

The correct answer is E.