Let S be a set of outcomes and let A and B be events with outcomes in

Question

Let S be a set of outcomes and let A and B be events with outcomes in S. Let ∼B denote the set of all outcomes in S that are not in B and let P(A) denote the probability that event A occurs. What is the value of P(A) ?

(1) P(A ⋃ B) = 0.7
(2) P(A ⋃∼B) = 0.9

DS95491.01

(1) P(A ⋃ B) = 0.7
(2) P(A ⋃∼B) = 0.9

DS95491.01

MahmoudFawzy · Accepted Answer

let A(only) + B(only) + AB(only) + neither = 1

from statement (1), it is clearly insufficient but it it means that neither = 0.3
from statement (2), it is clearly insufficient but it it means that B(only) = 0.1

by combining them, P(A) = A(only) + AB(only) = 1- (0.1+0.3) = 0.6 --->  sufficient 
 C

BrentGMATPrepNow · Answer

Target question :    What is the value of P(A) ?  
This is a good candidate to use the  Double Matrix method 
We can set up our matrix as follows: 
 https://i.imgur.com/fDlitvt.png 
Noticed that I've let 100 = the total number of possible outcomes. 
In order to determine the value of  P(A) , we need to find the total number of outcomes (out of 100) in which event A occurs. 
In other words, we need to find the sum of the top two boxes.

Statement 1 : P(A ⋃ B) = 0.7  
In other words, P(A or B) = 0.7
(0.7)(100) = 70
So, there is a total of 70 outcomes in which events A or B (or both) occur. 
In other words, we know that the sum of the three highlighted boxes below must add to 70. 
 https://i.imgur.com/Yuy3yIp.png 
Since all four boxes must add to 100, we know that the last (unshaded) box must be 30
Since the target question requires us to find that the sum of the top two boxes, statement 1 is NOT SUFFICIENT

Statement 2 : P(A ⋃∼B) = 0.9 
In other words, P(A or ~B) = 0.9
(0.9)(100) = 90
So, there is a total of 90 outcomes in which events A or ~B (or both) occur. 
In other words, we know that the sum of the three highlighted boxes below must add to 90. 
 https://i.imgur.com/k4WPjYZ.png 
Since all four boxes must add to 100, we know that the last (unshaded) box must be 10
Since the target question requires us to find that the sum of the top two boxes, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined    
When we combine both statements we get the following: 
 https://i.imgur.com/A3c2MnY.png 
Since the two bottom of boxes add to 40, we we know that P(~A) = 40/100 = 0.4
Since all four boxes must add to 100, we know that the TOP two boxes must add to   60  , which means   P(A) = 60/100 = 0.6  
Since we can answer the  target question  with certainty, the combined statements are SUFFICIENT

Answer: C

Cheers,
Brent

RELATED VIDEO
 https://www.youtube.com/watch?v=jK-tiBrrf04

menonpriyanka92 · Answer

Here is my approach

Information gathering:-
Please refer to the image attached. From the image:
x - p(A)  ONLY 
z - p(B)  ONLY 
y - p(A and B)
w - p(Neither A nor B) 
From the question, we know that set S is the total outcome. Hence w+x+y+z = 1...(Info.1)
As per the Q-stem, P(~B) = w+x

To find: P(A) = x+y = ?

Statement 1: P(A ⋃ B) = 0.7
means x+y+z = 0.7
We can conclude that w = 0.3 (1-x-y-z...refer to Info.1)
We need to know z as well to find out x+y..hence  insufficient.

Statement 2: P(A ⋃∼B) = 0.9
We know that P(~B) = w+x
P(A) = x+y 
So w+x+y = 0.9..(note this won't be w+2x+y because it is A U B - A union B....Please refer to venn diagram property if you get stuck here.)
we can infer that z = 0.1..but to get x+y we need to know w as well.. Insufficient

(1) and (2) together
We have w as well as z. substitute this in Info. 1 to find x+y... Sufficient

hspruthi76 · Answer

P(A U ~B) = 0.9 => P(B only) = 1-0.9 = 0.1
Now P(A U B) = P(A) + P(B only) => P(A) = 0.7-0.1 = 0.6

Kritisood · Answer

P(A) + P(B) = 0.7 (st 1)
P(A) + P('B) = 0.9 (st 2)
subtracting 1 and 2

P(B)-P('B)= -0.2
P(B)-  = -0.2
2P(B)=0.8
P(B)=0.4 putting this value in statement 1 gives us the result.  answer C

nick1816   chetan2u  even though I'm getting the right answer, my answer for P(A) is coming as 0.3. could you please help with where I am faltering?

nick1816 · Answer

You assumed that the 2 events A and B are mutually exclusive, which means events A and B don't have common outcomes or P(A⋂B) = 0. (In statement 1). Untitled.png In statement 2, you assumed A and ~B are mutually exclusive or P(A⋂~B)=0 Untitled.png Your assumptions in both statements are contradicting. Hence, you're getting wrong values.

reynaldreni · Answer

In general we know, A+X+B+Z = 1 
We need the value of A+X = ?

(1) P(AUB)=0.7
This implies A+X+B = 0.7 
We cannot find A+X from the above >>> Insuff.

(2) P(AU~B)=0.9
This implies A+X+Z = 0.9
From this we cannot deduce the value of A+X >>> Insuff.

(1)+(2)
A+X+B+Z = 1 --(p)
A+X+Z = 0.9 --(q)
A+X+B = 0.7 --(r)

Eq, (p)-(q) We get, B=0.1
using Eq, (r) A+X+0.1 = 0.7
Therefore, A+X = 0.6

OA C

suminha · Answer

IMO C.

Since the common denominator is 10, we can let S be 10.

Posted from my mobile device

chetan2u · Answer

Hi  dave13

B is part of S, and ~B is everything else in S, that is B+(~B)=Set S

For example 
Say S is the set of all rational numbers(integers +fraction) and B is set of even numbers, then ~B is odd numbers+fractions(with denominator other than 1)

Or say S is set of integers, while B is the outcomes when you draw a dice, that is 1,2,3,4,5,6.
So ~B is ....-3,-2,-1,0,7,8,9....

sambitspm · Answer

I was confused between (A Union Not B) and (A Intersection Not B) and that was the reason why I got this question wrong twice 😁. Here is the screenshot of the Union and Intersection. I hope this clears the air.

UPDATE:
Thanks  Poojita  for pointing,

The screenshot is related to concept where there is  no possibility of none of them occurring.  To clarify this I am attaching my solution to this problem

GMATinsight · Answer

Answer: Option C

Video solution by  GMATinsight

https://www.youtube.com/watch?v=TPrSkfXB9jE

frankgraves · Answer

I got C without any Venn diagram.
This is how I did it.

One important thing to remember :  P(B) + P(~B) = 1 
Because S is a set of outcomes, the total accumulation of all probabilities in S must be 1.
So if we add P(~B), which is all outcomes outside of B, with P(B), we are basically adding all elements in S, which equals to 1.
Is there any element in S besides A and B? Is there any overlap between A and B? We do not know and we do not need to know.
Now let's go to the statements.

(1) 
P(A ⋃ B) = 0.7
P(A) + P(B) = 0.7 (i)
Insufficient

(2) 
P(A ⋃ ~B) = 0.9
P(A) + P(~B) = 0.9 (ii)
Insufficient

(1) + (2)
Adding equation (i) with equation (ii)
2P(A) + P(B) + P(~B) = 1.6
 Remember that P(B) + P(~B) = 1 
2P(A) + 1 = 1.6
2P(A) = 0.6
P(A) = 0.3
Sufficient
Answer is C

dave13 · Answer

VeritasKarishma re: St1 if P(A ∪ B) means probability of A OR B then we should have this P(A ∪ B)= P(A) + P(B) - P(A ∩ B) why in the posts above they add BOTH ?

Also Basing solely on info below (without reading statements 1 and 2) can you figure out if events are mutually exclusive or independent ? if yes how ? "Let S be a set of outcomes and let A and B be events with outcomes in S. Let ∼B denote the set of all outcomes in S that are not in B and let P(A) denote the probability that event A occurs. What is the value of P(A) ?"

KarishmaB · Answer

Consider the set S (the rectangle). A and B are subsets in it. They may have some overlap. We need A which is shown by yellow + green. Screenshot 2021-06-25 at 10.18.44.png (1) P(A ⋃ B) = 0.7 This means the yellow + green + blue = 0.7 We need yellow + green. Not sufficient (2) P(A ⋃∼B) = 0.9 This means yellow + green + white = 0.9 (because we need any element which is either in A (so all elements of A work) or not in B (all white elements) or both (it is in A but not in B). We need yellow + green. Not sufficient. Using both, yellow + green + blue + yellow + green +white = 0.7 + 0.9 = 1.6 Now note that yellow + green + blue + white = 1 So yellow + green + 1 = 1.6 yellow + green = 0.6 Sufficient Answer (C) Also Check: https://youtu.be/HRnuURqGhmg Bambi2021

KarishmaB · Answer

You are right. 
P(A U B) = P(A) + P(B) - P(Both)

No, the sets are not given to be mutually exclusive and you don't have to worry about whether they are. In the most general case, just assume that there would be some overlap. 
Also, whether the events are independent or not is irrelevant. Note that independent events can have an overlap (overlap means that both happened). For example, if I say that "it will rain today (Probability of 0.4)" and "My team will win its match today (Probability of 0.6)" are independent events, it just means that one is not dependent on another. It is possible that both things happen today. Probability of that is 0.4*0.6 = 0.24.

Don't let ∼B bother you. Just identify the region on your venn diagram and proceed as usual.

gmatbyexample · Answer

S is set of all possible outcomes, A and B are subsets. 
P(A) = A/S, P(B) = B/S and P(S) = S/S = 1

Also, P(B) + P(~B) = P(S) = 1 (since ~B has all elements in S, which are not in B).

Also, P(AnB) + P(An~B) = P(A) --> (See video for analysis on this one)

With these, it is easy to follow the analysis into finding P(A) = 0.6

0.9=P(AUB) = P(A) + P(B) - P(AnB)
0.7=P(AU~B) = P(A) + P(~B) - P(An~B)
1.6 = 2P(A) + (P(B) + P(~B)) - ( P(AnB) + P(An~B) )
1.6 = 2P(A) + 1 - P(A)
=> P(A) = 0.6

https://www.youtube.com/watch?v=e3bk1hoL7hY

PyjamaScientist · Answer

Credit to Sanjitscorps18 for this answer. He sent it to my PM. But, this is so nicely written that it should be here and in PM.

S consists of {A, B, any other outcomes} Now if we visualize this from a Venn diagram perspective, there is a possibility that A and B can have an overlap. Now using this in eq(1) we get S (Total outcomes) = Outcomes (A) + Outcomes (B) + Outcomes (A and B) + Outcomes (Others) S (Total outcomes) = Outcomes (A) + Outcomes (B) + Outcomes (A and B) + Outcomes (Neither A nor B) When we take probability of the above equation we get S (Total Probability) = Probability(A) + Probability(B) + Probability(A and B) + Probability(Neither A nor B) => 1 = P(A) + P(B) + P(A and B) + P(neither) ----- (1) Essentially we need the value of P(A) + P(A and B) as the question asks what is the probability of A Now taking the individual statements 1. P(AUB)=0.7 P(A) + P(B) + P(A and B) = 0.7 => P(neither) = 0.3 (2) P(AU~B)=0.9 P(A) + P(A and B) + P(neither) = 0.9 => P(B) = 0.1 Clearly, we cannot find P(A) + P(A and B) from either of the statements alone Substituting the values in equation (1) above we get 1 = P(A) + P(B) + P(A and B) + P(neither) => 1 = P(A) + 0.1 + P(A and B) + 0.3 => P(A) + P(A and B) = 1 - 0.3 - 0.1 = 0.6 => P(A) + P(A and B) = 0.6

AutumnChen · Answer

"Let ∼B denote the set of all outcomes in S that are not in B"

Why some of your explanations said P(A ⋃∼B) = P(only A) + both + P (neither)?

Why not P(A ⋃∼B) = P(only A) + P (neither)?

nikita.nanda1810 · Answer

Hi!  Bunuel  Can you please help explain by P (A U ~B) -- P (only A) + P (A&B) + P(neither)

Why does this include P A&B) given that ~B is all outcomes without B?

I marked incorrectly because I understood P(A U ~B) = P (only A) + P(Neither)

Any clarity would help.

Thanks! 
Nikita

Let S be a set of outcomes and let A and B be events with outcomes in

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