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President's chances of correct decision

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President's chances of correct decision

by S0laris » Sun Apr 05, 2009 10:45 am
President has 3 advisors. 1st-adviser mistakes 10% times. 2nd-adviser mistakes 20% times. 3rd-adviser mistakes 30% times.
Each adviser gives with 1 advise to every one president's decision. President makes final decision basing on all given advises.
If there is only one decision to make, what are the chances of president's mistake ?

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it's not the test, so there is no answer choices to guess, just straight methods to solve. I'm interested in solving options. Answer after some discussions.
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Re: President's chances of correct decision

by dtweah » Sun Apr 05, 2009 4:15 pm
S0laris wrote:President has 3 advisors. 1st-adviser mistakes 10% times. 2nd-adviser mistakes 20% times. 3rd-adviser mistakes 30% times.
Each adviser gives with 1 advise to every one president's decision. President makes final decision basing on all given advises.
If there is only one decision to make, what are the chances of president's mistake ?

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it's not the test, so there is no answer choices to guess, just straight methods to solve. I'm interested in solving options. Answer after some discussions.
Let M1 be the event that Advisor one makes a mistake
M2 event of Advisor 2's mistake
M3 even of advisor 3's mistake

then Probability of mistake is given as

P(M1)=1/10
P(M2)=1/5
P(M3) 3/10

If we let E be the event that the president makes a mistake, then since one advisor's probabity of making mistake is independent of another's and the president has to listen to all three:

P(E)= 1/10 x 1/5 x 3/10 = 3/500.

A more interesting question is what is the president's chance of mistake given at most two advisors make a mistake? I leave that for discussion.
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Re: President's chances of correct decision

by S0laris » Sun Apr 05, 2009 11:34 pm
dtweah wrote: Let M1 be the event that Advisor one makes a mistake
M2 event of Advisor 2's mistake
M3 even of advisor 3's mistake

then Probability of mistake is given as

P(M1)=1/10
P(M2)=1/5
P(M3) 3/10

If we let E be the event that the president makes a mistake, then since one advisor's probabity of making mistake is independent of another's and the president has to listen to all three:

P(E)= 1/10 x 1/5 x 3/10 = 3/500.

A more interesting question is what is the president's chance of mistake given at most two advisors make a mistake? I leave that for discussion.
You considered only 1 way that situation may be developed - all 3 advisers give incorrect advises. What about the other chances, such as 1-correct, 2-incorrect, and 3-correct or 1-incorrect, 2-correct, and 3-incorrect and so forth ?
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by scoobydooby » Mon Apr 06, 2009 5:17 am
prob( Presidents mistake)= 1-prob(all ministers correct) (mutually exclusive events)

prob(1st minister correct)=90%
prob (2nd minister correct)=80%
prob (3rd minister correct)= 70%

prob (Presidents mistake)= 1-(90/100*80/100*70/100)
= 1-(9/10*4/5*7/10)
=248/500
=62/125
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by S0laris » Mon Apr 06, 2009 5:27 am
scoobydooby wrote:prob( Presidents mistake)= 1-prob(all ministers correct) (mutually exclusive events)

prob(1st minister correct)=90%
prob (2nd minister correct)=80%
prob (3rd minister correct)= 70%

prob (Presidents mistake)= 1-(90/100*80/100*70/100)
= 1-(9/10*4/5*7/10)
=248/500
=62/125
62/125=0.496 or 49% - seems like president mistakes more frequently than his advisers do ?
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by scoobydooby » Mon Apr 06, 2009 5:49 am
yes, guess dictatorship would serve him better. :)
the more the number of ministers, the more the president will have to take all their advices into consideration, more his chances of mistake.

the probabilities are fractions, as we keep multiplying them, the product becomes smaller and smaller
so prob(President mistake):1-prob(all ministers correct) actually increases.
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by Vemuri » Mon Apr 06, 2009 7:05 am
scoobydooby wrote:prob( Presidents mistake)= 1-prob(all ministers correct) (mutually exclusive events)

prob(1st minister correct)=90%
prob (2nd minister correct)=80%
prob (3rd minister correct)= 70%

prob (Presidents mistake)= 1-(90/100*80/100*70/100)
= 1-(9/10*4/5*7/10)
=248/500
=62/125
Hi scoobydooby. What is the difference between:

The probability that the president makes mistake = probability that all the advisors are wrong.

vs

The probability that the president makes mistake = 1- (probability that all the advisors are correct).

Both the answers should give the same probability. Is this a correct assumption or am I missing a concept here?
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by scoobydooby » Mon Apr 06, 2009 8:02 am
hi vemuri,

my guess was that if even one of the ministers made a mistake, the president would make a mistake. this could have various combinations, for eg. 1st wrong, 2nd correct, 3rd correct; 1st wrong, 2nd wrong, 3rd incorrect, all wrong...........etc, etc.
so i elimated the possibility of all being correct from 1 which would take care of all the above possible cases of mistakes.

however, am not too sure how exactly is a mistake by the president defined- whether for a mistake by the president all the ministers need necessarily make a mistake or even 1 mistake by a minister would suffice.

may be i am missing. will wait for the OA/OE and other approcahes :)
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by kapsii » Mon Apr 06, 2009 9:50 am
a lame attempt at the question:

for say 100 topics of discussion, the president can get up to 300 advices in total from his advisers, out of which (10 + 20 + 30 = 60) will probably be wrong.

Since he has to choose from one of them (and he is seemingly totally dependent on his advisers) his probability of getting the decision wrong is : 60/300 or 1/5 or 0.20.

assumptions made:
All his advisers give different advise, they never agree (seems realistic!)
President will choose those decisions with out using his wisdom, i.e. there are no other external factors.

It makes sense to me, as the question has no pointers on how the decision would be made, if the advisers have different weightage, thus we can assume that every one is an equal and every advise has equal probability of turning into decision.
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