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Intern
Joined: 18 Jul 2010
Posts: 45
Followers: 0

Kudos [?]: 10 [0], given: 6

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Question Stats:

100% (02:53) correct 0% (00:00) wrong based on 1 sessions
Hi everyone !
Could you explain me the answer of the question?
[Reveal] Spoiler: OA

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GMAT Club team member
Joined: 02 Sep 2009
Posts: 12115
Followers: 1879

Kudos [?]: 10120 [1] , given: 961

1
KUDOS
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.

_________________
Intern
Joined: 18 Jul 2010
Posts: 45
Followers: 0

Kudos [?]: 10 [0], given: 6

Like the second option.
You are damn fast!
Thanks a lot!
Manager
Joined: 01 Nov 2010
Posts: 184
Location: Zürich, Switzerland
Followers: 2

Kudos [?]: 11 [0], given: 20

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

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