Problem Solving

Hi All,

We're told that 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 does not to sign Bill X, she will sign either Bill Y or Bill Z, chosen randomly. The 'intent' of this question is to ask for the percentage chance that the candidate will sign Bill Z.

To start, this is an oddly-worded question, but it's essentially a big Probability question (with each additional 'step' decreasing the probability of the end result). For Bill Z to get signed, a number of events have to occur first:
-This one candidate must win the election. (40% chance)
-The candidate must NOT sign Bill X. (40% chance)
-Between Bill Y and Bill Z, Bill Z must be the one that's randomly chosen (50% chance)

The probability of all of these events occurring is (.4)(.4)(.5) = .080 = 8%

Final Answer: B

GMAT assassins aren't born, they're made,
Rich

BTGmoderatorLU 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 Bill X, she will sign either Bill Y or Bill Z, chosen randomly. What is the chance that the candidate will sign Bill Z?

A. 10
B. 8
C. 6
D. 4
E. 5

The candidate will win with 40% = 2/5 probability.

After she wins, she will sign Bill X with 60% = 3/5 probability; therefore, she has a 40% = 2/5 chance of signing other bills. Since she will sign either Bill Y or Bill Z with an equal chance, she has a 50% = 1/2 probability of signing Bill Z. Therefore, if she is elected, the probability she will sign Bill Z is

2/5 x 2/5 x 1/2 = 2/25 = 8/100 = 8%.

Answer: B