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pareekpuneet
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For everyone's benefit
Q1
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Q2
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Q2)
Prob of Leo to hear a song he likes =
Prob of Radio A playing songs which Leo likes
+Prob of Radio B playing songs which Leo likes
+Prob of Radio C playing songs which Leo likes
= 0.3 + 0.3 * 0.7 +0.3 *0.3 *0.7
= 0.3+0.21+0.147
=0.657
Option D
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Time spent with 5 mile@30mph = X-5/60 + 5/30 = (X+5)/60
Time spent with entire trip @60mph = X/60.
Time Diff = 1/12
Q: % of diff with total time @60mph:
(1/12) / (X/60)*100
= (500/X)%
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Hi Guys,

Time distance was so simple... i just screwed it ... dont know y ... thanks for that one .. but for the probability one can anyone please elaborate the below mentioned logic for me as i dont know much about the proability:

1. Radio A --> 0.3 (I understand)
2. Radio B --> 0.3 (For radio B) * 0.7 (for the case where radio A is not played)
3. Radio C --> 0.3 (For radio C) * 0.3 (Why this 0.3 and not 1 - 0.3 = 0.7) * 0.7.

I dont understand the logi behind two times 0.3 in the last case as mentioned by irajeevsingh.

i know it might b simple... not clicking to me :( :( ....

Thanks for your help.

Regards,
Aban
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Look at this file. This problem is a decision tree problem.



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