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# 1-The probability of passing test A is a, The probability of

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CEO
Joined: 15 Aug 2003
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1-The probability of passing test A is a, The probability of [#permalink]  07 Dec 2003, 21:39
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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
Director
Joined: 28 Oct 2003
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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.
Intern
Joined: 20 Oct 2003
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I would vote for #3 as both events are independent,
thus p(a)+p(b)- p(a)*p(b).

Racer.
CEO
Joined: 15 Aug 2003
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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.
Intern
Joined: 20 Oct 2003
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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.
Director
Joined: 13 Nov 2003
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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
CEO
Joined: 15 Aug 2003
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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).
CEO
Joined: 15 Aug 2003
Posts: 3469
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Kudos [?]: 701 [0], given: 781

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)..
Director
Joined: 28 Oct 2003
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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.
Director
Joined: 13 Nov 2003
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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
SVP
Joined: 03 Feb 2003
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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
Intern
Joined: 18 Sep 2003
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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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