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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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Question Stats:
71% (02:09) correct
29% (01:19) wrong based on 223 sessions
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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? A. 0.027 B. 0.090 C. 0.417 D. 0.657 E. 0.900 OPEN DISCUSSION OF THIS QUESTION IS HERE: onhisdrivetoworkleolistenstooneofthreeradiostat104659.html
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Re: On his drive to work, Leo listens to one of three radio [#permalink]
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11 Oct 2007, 10:01
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assumption: if he hears a song he likes, he keeps listening to that station:
solution= 0.3 +0.7x0.3+0.7x0.7x0.3 =0.657



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Re: On his drive to work, Leo listens to one of three radio [#permalink]
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11 Oct 2007, 12:59
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1st scenario:
picking A (then he wont change stations anymore)
0.3
2nd scenario:
picking B (no A, pick B, then he stops)
0.7 * 0.3 = .21
3rd scenario:
picking C (no A, no B, pick C)
0.7 * 0.7 * 0.3 = .147
adding the three
0.3 + .21 + .147 = 0.657



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Re: On his drive to work, Leo listens to one of three radio [#permalink]
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11 Oct 2007, 13:52
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ronron wrote: assumption: if he hears a song he likes, he keeps listening to that station:
solution= 0.3 +0.7x0.3+0.7x0.7x0.3 =0.657
OR
We can use P(A) = 1  P(A')
P(Song He likes) = 1  P(Song He does not like at all)
P(Song He likes) = 1  (0.7 x 0.7 x 0.7)
= 1  0.343
= 0.657



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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
Total = 0.657
Ans D



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Re: On his drive to work, Leo listens to one of three radio [#permalink]
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29 Dec 2014, 23:33
Very good question. Please Solve this. Bumping it up
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Re: On his drive to work, Leo listens to one of three radio [#permalink]
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30 Dec 2014, 01:56
leeye84 wrote: 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: 1the probability that neither of the stations is playing the song he likes: \(P=10.7*0.7*0.7=0.657\). Answer: D. OPEN DISCUSSION OF THIS QUESTION IS HERE: onhisdrivetoworkleolistenstooneofthreeradiostat104659.html
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Re: On his drive to work, Leo listens to one of three radio
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