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

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11 Nov 2010, 08:00

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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?

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: \(P=1-0.7*0.7*0.7=0.657\).

On a station, the probability of getting song of choice is - 3/10. The probability of not getting the song of choice on any station therefore is - 7/10.

So, Probability of getting a song of choice on at least one of the stations

= 1-Probability of not getting a song of choice while trying all 3 stations. = 1 - (7/10)*(7/10)*(7/10) = 1- 343/1000 = 657/1000 =.657

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

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05 Feb 2016, 11:06

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