Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 24 May 2017, 20:53

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
CEO
Joined: 15 Aug 2003
Posts: 3454
Followers: 67

Kudos [?]: 874 [0], given: 781

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

### Show Tags

07 Dec 2003, 22:39
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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
Posts: 501
Location: 55405
Followers: 1

Kudos [?]: 25 [0], given: 0

### Show Tags

07 Dec 2003, 22:45
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
Posts: 13
Location: Moscow
Followers: 0

Kudos [?]: 0 [0], given: 0

### Show Tags

07 Dec 2003, 23:58
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
Posts: 3454
Followers: 67

Kudos [?]: 874 [0], given: 781

### Show Tags

08 Dec 2003, 00:56
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
Posts: 13
Location: Moscow
Followers: 0

Kudos [?]: 0 [0], given: 0

### Show Tags

08 Dec 2003, 01:35
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
Posts: 789
Location: BULGARIA
Followers: 1

Kudos [?]: 52 [0], given: 0

### Show Tags

08 Dec 2003, 02:50
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
Posts: 3454
Followers: 67

Kudos [?]: 874 [0], given: 781

### Show Tags

08 Dec 2003, 03:06
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: 3454
Followers: 67

Kudos [?]: 874 [0], given: 781

### Show Tags

08 Dec 2003, 03:12
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
Posts: 501
Location: 55405
Followers: 1

Kudos [?]: 25 [0], given: 0

### Show Tags

08 Dec 2003, 07:39
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
Posts: 789
Location: BULGARIA
Followers: 1

Kudos [?]: 52 [0], given: 0

### Show Tags

08 Dec 2003, 09:29
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
Posts: 1604
Followers: 9

Kudos [?]: 268 [0], given: 0

### Show Tags

09 Dec 2003, 04:43
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
Posts: 2
Location: Chennai
Followers: 0

Kudos [?]: 0 [0], given: 0

### Show Tags

09 Dec 2003, 04:55
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
09 Dec 2003, 04:55
Display posts from previous: Sort by