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On his drive to work, Leo listens to one of three radio [#permalink]

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11 Oct 2007, 09:40

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On his drive to work, Leo listens to one of three radio stations A, B or C. He first turns to A. If A is playing a song he likes, he listens to it; if not, he turns it to B. If B is playing a song he likes, he listens to it; if not, he turns it to C. If C is playing a song he likes, he listens to it; if not, he turns off the radio. For each station, the probability is 0.30 that at any given moment the station is playing a song Leo likes. On his drive to work, what is the probability that Leo will hear a song he likes?

Re: On his drive to work, Leo listens to one of three radio [#permalink]

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11 Oct 2007, 23:26

Probability he hears a song is like:

Tune to A and likes what he is hearing = 0.3
Tune to A, don't find he like what they are airing, then tune to B and likes what he finds there = 0.7 * 0.3 = 0.21
Tune to A, finds crap there, Tune to B, hears a similar crap, Tune to C and finally falls in love with the program = 0.7^2 * 0.3 = 0.147

On his drive to work, Leo listens to one of three radio stations A, B or C. He first turns to A. If A is playing a song he likes, he listens to it; if not, he turns it to B. If B is playing a song he likes, he listens to it; if not, he turns it to C. If C is playing a song he likes, he listens to it; if not, he turns off the radio. For each station, the probability is 0.30 that at any given moment the station is playing a song Leo likes. On his drive to work, what is the probability that Leo will hear a song he likes?

A. 0.027 B. 0.090 C. 0.417 D. 0.657 E. 0.900

The desired probability is the sum of the following events:

A is playing the song he likes - 0.3; A is not, but B is - 0.7*0.3=0.21; A is not, B is not, but C is - 0.7*0.7*0.3=0.147; \(P=0.3+0.21+0.147=0.657\).

Or: 1-the probability that neither of the stations is playing the song he likes: \(P=1-0.7*0.7*0.7=0.657\).

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