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

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Senior Manager
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On his drive to work, Leo listens to one of three radio [#permalink] New post 11 Oct 2007, 08:40
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A
B
C
D
E

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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 plays a song he likes, he listens to it. If not, he turns to B. If B is playing a song he likes, he listens to it. If not, he turns 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?

1)0.027
2)0.090
3)0.417
4)0.657
5)0.900
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 [#permalink] New post 11 Oct 2007, 09:01
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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 [#permalink] New post 11 Oct 2007, 11:59
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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 [#permalink] New post 11 Oct 2007, 12:52
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
GMAT Club Legend
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 [#permalink] New post 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

Ans D
  [#permalink] 11 Oct 2007, 22:26
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On his drive to work, Leo listens to one of three radio

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