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CEO
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1-The probability of passing test A is a, The probability of [#permalink]
 07 Dec 2003, 22:39
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0% (00:00) correct
 0% (00:00) wrong based on 0 sessions
1-The probability of passing test A is a, The probability of
passing test B is b, if a student participates in both tests what is
the probability of passing either A or B but not both ?
1. a+b 2.a+b-2ab 3.a+b-ab 4.(1-a)*(1-b) 5.(a-b)(b-a)
kinda simple..just want to see how you guys solve it
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Director
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I didn't know the formula, but I plugged a few numbers and got #2.
I'd like to see someone provide a little theoretical backing. I worry that my "pick some numbers and plug" strategy will backfire in a high stress situation.
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Intern
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I would vote for #3 as both events are independent,
thus p(a)+p(b)- p(a)*p(b).
Racer.
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CEO
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Racer wrote: I would vote for #3 as both events are independent,
thus p(a)+p(b)- p(a)*p(b).
Racer.
No. There is no information about events being independent.
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Intern
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Well, my logic was:
1. the probability of either of events occuring is a sum of their probabilities:
a+b
2. the tasks says: 'but not passing both' means we have to deduct the possibility of passing both tests i.e. p(a)+p(b)-p(a*b)
Though, if we can assume that passing test A automaticaly means not passing or no need to pass test B then Probability is simply a+b as p(a*b)=0
Appreciate yr feedback.
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Director
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Support Racer, P(a or b )=Pa+Pb-P(a and b), Example Pa=50%, Pb=70% P(A or B)=70%+50%-35%=85%, if it was A+B-2AB then it is 120%-70%=50% which makes no sense since the prob of passing either should be higher than the probability of passing a single exam...IMO 
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CEO
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Draw a tree diagram , the problem would be easier to see.
Prob of Passing only A = Prob of success in A * Prob of failure in B
= a * (1-b) = a - ab
Similarly, Prob of Passing only B = b * (1-a) = b - ba
The probability of either A or B = a - ab + b - ba = a +b -2ab
its choice (2).
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CEO
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racer wrote:
Quote:
Well, my logic was:
1. the probability of either of events occuring is a sum of their probabilities: a+b
2. the tasks says: 'but not passing both' means we have to deduct the possibility of passing both tests i.e. p(a)+p(b)-p(a*b)
You use the wrong concept here.
a => prob of passing
b => prob of passing
a*b => prob of passing both
When you use the "union formula" , you are calculating the probability of passing both.
Nowhere in your solution have you included the probability of failure of a ...which is (1-a) ..and b which is (1-b)..
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Director
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Racer and BG, consider this:
Equation 3 is a+b-ab
Say there's a 100% chance of passing test a, and a 100% chance of passing test b.
1+1-(1*1) =1
The question asks for "the probability of passing either A or B but not both?". The answer to this question is zero, if the chance of passing each test is 1.
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Director
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Stolfi, accept your solution, you are right that the answer should be a+b-2ab since the ab part should be completely ommited...thanks
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SVP
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since there is info that the events are independent, let us go straight.
P=a(not b)+b(not a)=a(1–b)+b(1–a)=a+b–2ab
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Intern
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P(AuB) = P(A) + P(B) - P(A)P(B), as the events are independent
Not both means we have to subtract the probability of passing both.
P(AuB) = P(A) + P(B) - P(A)P(B) - P(A)(B)
= a+b-ab-ab = a+b-2ab
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