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

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On his drive to work, Leo listens to one of three radio [#permalink]  11 Oct 2007, 08:40
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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?

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

OPEN DISCUSSION OF THIS QUESTION IS HERE: on-his-drive-to-work-leo-listens-to-one-of-three-radio-stat-104659.html
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Re: On his drive to work, Leo listens to one of three radio [#permalink]  11 Oct 2007, 09:01
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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]  11 Oct 2007, 11:59
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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

0.3 + .21 + .147 = 0.657
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Re: On his drive to work, Leo listens to one of three radio [#permalink]  11 Oct 2007, 12:52
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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]  11 Oct 2007, 22:26
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

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Re: On his drive to work, Leo listens to one of three radio [#permalink]  29 Dec 2014, 22:33
Expert's post
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]  30 Dec 2014, 00:56
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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: 1-the probability that neither of the stations is playing the song he likes: $$P=1-0.7*0.7*0.7=0.657$$.

OPEN DISCUSSION OF THIS QUESTION IS HERE: on-his-drive-to-work-leo-listens-to-one-of-three-radio-stat-104659.html
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Re: On his drive to work, Leo listens to one of three radio   [#permalink] 30 Dec 2014, 00:56
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