Probability Problem Set 2

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theachiever: Master | Next Rank: 500 Posts; Posts: 116; Joined: Wed Oct 03, 2012 10:44 pmFollowed by: 1 member; Thanked: 5 times

by theachiever » Thu Nov 01, 2012 1:27 am

A polling company reports that there is a 40% chance that a certain candidate will win the next election.If the candidate wins,there is a 60% chance that she will sign Bill X and no other bills.
If she decides not to sign Bil X,she will sign either Bill Y or Bill Z,chosen randomly.What is the chance that the candidate will sign Bill Z?

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Source: Beat The GMAT — Problem Solving | Thu Nov 01, 2012 1:27 am

pemdas: Legendary Member; Posts: 1084; Joined: Fri Apr 15, 2011 2:33 pmFollowed by: 21 members; Thanked: 158 times

by pemdas » Thu Nov 01, 2012 3:17 am

theachiever wrote:A polling company reports that there is a 40% chance that a certain candidate will win the next election.If the candidate wins,there is a 60% chance that she will sign Bill X and no other bills.
If she decides not to sign Bil X,she will sign either Bill Y or Bill Z,chosen randomly.What is the chance that the candidate will sign Bill Z?

According to condition a candidate will win but refuse to sign Bill X, that is 40% chance of candidate's win combined with (100-60)% chance of not signing Bill X, i.e. singing either Bill Y or Z.

0,4*(1-0,6)=p(Bill Y) + p(Bill Z). Since we consider randomness in selection of Bills Y or Z, that is no special preference between two Bills and either one can be selected, there is 50/50 % chance each will be selected. Hence p(Bill Z)=p(Bill Y) Or 0,4*0,4=p(Bill Z) + p(Bill Z), 0,16=2*p(Bill Z) and p(Bill Z)=0,16/2=0,08

chance that the candidate will sign Bill Z is 0,08 or 8%

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