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twobagels
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Does someone have the answer for this?

I couldn't quite solve it even using a table approach

P(A and B) = x
P(B and Not A) = 1.5y
P(A and Not B) = y
P(Neither) = 6x

Got stuck after.
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can someone explain the answer of this
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Here is my approach
Remember: Events A and B are independent then P(A∩B) = P(A) · P(B)
Note: P(A) is the probability of event A occur
P(A') is the probability of event A NOT occur (1-P(A)
P(B) is the probability of event B occur
P(B') is the probability of event B NOT occur (1-P(B)
Having: A and B occur independently of one another (A occur not depend on B)=> P(A and B) = P(A)*P(B)
=> A' and B' occur independently (or P(A' and B') = P(A')*P(B'))
A and B' occur independently (or P(A and B') = P(A)*P(B'))
B and A' occur independently (or P(B and A') = P(B)*P(A'))
On the question:
1. The probability that neither event occurs (P(A' and B')) is 6 times the probability that both events occur(P(A and B)).
\(=> P(A' and B') = 6*P(A and B)\\
<=>P(A')*P(B') = 6*P(A)*P(B) (1)\)
2. the probability that event B occurs and event A does not occur(P(B and A')) is 50 percent greater than the probability that event A occurs but event B does not occur (P(A and B'))
=> P(B and A') is 50% greater than P(A and B')
So if P(A and B') is 100% then P(B and A') is 100%+50% = 150% P(A and B')
=> P(B and A') = 150% P(A and B')
<=> P(B)*P(A') = 150%P(A)*P(B') (2)
From (1) (2), divide (1) by (2)
\(\frac{P(A')*P(B')}{(P(B)*P(A')} = \frac{6*P(A)*P(B)}{(150%*P(A)*P(B')}\)
\(<=> \frac{P(B')}{P(B)} = \frac{4*P(B)}{P(B')}\)
\(<=>\frac{P(B')^2}{P(B)^2} = 4\)
\(<=>\frac{P(B')}{P(B)} = 2\)
<=> P(B') = 2P(B)
Having P(B)+P(B') = 1
=> P(B)+2P(B) = 1
=> 3P(B) = 1
=> P(B) = 1/3
Hence answer C
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chetan2u
twobagels
Events A and B occur independently of one another. The probability that neither event occurs is 6 times the probability that both events occur. Furthermore, the probability that event B occurs and event A does not occur is 50 percent greater than the probability that event A occurs but event B does not occur. What is the probability that event B occurs?

A. 1/5
B. 1/4
C. 1/3
D. 1/2
E. 3/4


Let the probability that event A and B occur be A and B respectively, so probability that A and B do not occur are (1-A) and (1-B) respectively.

Quote:
\(3B^2+3B-B-1=0.......3B(B+1)-1(B+1)=0.......(3B-1)(B+1)=0\)

So B=-1 or B=\(\frac{1}{3}\)

As B cannot be less than 0, B=\(\frac{1}{3}\)


C

The coeff of x^2 needs to be 1 for factorizing like that.
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chetan2u
twobagels
Events A and B occur independently of one another. The probability that neither event occurs is 6 times the probability that both events occur. Furthermore, the probability that event B occurs and event A does not occur is 50 percent greater than the probability that event A occurs but event B does not occur. What is the probability that event B occurs?

A. 1/5
B. 1/4
C. 1/3
D. 1/2
E. 3/4


Let the probability that event A and B occur be A and B respectively, so probability that A and B do not occur are (1-A) and (1-B) respectively.

Quote:
\(3B^2+3B-B-1=0.......3B(B+1)-1(B+1)=0.......(3B-1)(B+1)=0\)

So B=-1 or B=\(\frac{1}{3}\)

As B cannot be less than 0, B=\(\frac{1}{3}\)


C

The coeff of x^2 needs to be 1 for factorizing like that.


No, it need not be.

You have to multiply the coefficient of x^2 and the constant( the term without x), and try to get the coefficient of x in terms of two factors of that product.
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chetan2u
twobagels
Events A and B occur independently of one another. The probability that neither event occurs is 6 times the probability that both events occur. Furthermore, the probability that event B occurs and event A does not occur is 50 percent greater than the probability that event A occurs but event B does not occur. What is the probability that event B occurs?

A. 1/5
B. 1/4
C. 1/3
D. 1/2
E. 3/4


Let the probability that event A and B occur be A and B respectively, so probability that A and B do not occur are (1-A) and (1-B) respectively.

Quote:
The probability that neither event occurs is 6 times the probability that both events occur.
The probability that neither event occurs = (1-A)(1-B)
The probability that both events occur = A*B
So (1-A)(1-B)=6AB......1-A-B+AB=6AB
Since we are looking at value of B, let us get the equation in terms of B lone.
1-B=5AB+A.......\(A=\frac{1-B}{5B+1}\)...(i)

Quote:
Furthermore, the probability that event B occurs and event A does not occur is 50 percent greater than the probability that event A occurs but event B does not occur.
The probability that event B occurs and event A does not occur = B*(1-A)
The probability that event A occurs but event B does not occur = A(1-B)
So, \(B(1-A)=\frac{3}{2}*A(1-B)\)..........
\(2B-2AB=3A-3AB.......2B=3A-AB\).......
A=\(\frac{2B}{3-B}\)...(ii)

Equating (i) and (ii)
\(A=\frac{1-B}{5B+1}=\frac{2B}{3-B}\).........
\((1-B)(3-B)=2B(5B+1)\).........
\(3-4B+B^2=10B^2+2B\).......
\(9B^2+6B-3=0.........3B^2+2B-1=0\)
\(3B^2+3B-B-1=0.......3B(B+1)-1(B+1)=0.......(3B-1)(B+1)=0\)

So B=-1 or B=\(\frac{1}{3}\)

As B cannot be less than 0, B=\(\frac{1}{3}\)

C

chetan2u - you're a god. Even after having a 720, I'm going for my 6th attempt at the GMAT and in one single question you cleared up my concepts of probability, set theory, and word problems all at once. Thank you
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twobagels
Events A and B occur independently of one another. The probability that neither event occurs is 6 times the probability that both events occur. Furthermore, the probability that event B occurs and event A does not occur is 50 percent greater than the probability that event A occurs but event B does not occur. What is the probability that event B occurs?

A. 1/5
B. 1/4
C. 1/3
D. 1/2
E. 3/4

It's probably best to do the real math on this one, but just for anyone who gets stuck sometimes, we can also get there by Plugging In The Answers (PITA). I like trying B and D.

Let's start with B.
B = 1/4, so -B = 3/4
(1-A)(3/4)=6(A)(1/4)
(3-3A)/4=6A/4
3-3A=6A
3=9A
A=1/3, so -A = 2/3
What's the probability that B and -A? (1/4)(2/3) = 2/12 = 1/6
What's the probability that A and -B? (1/3)(3/4) = 1/4
Is 1/6 50% greater than 1/4? No, it's less than 1/4.
B is wrong.

Let's try D.
B = 1/2, so -B = 1/2
(1-A)(1/2)=6(A)(1/2)
(1-A)/2=6A/2
1-A=6A
1=7A
A=1/7, so -A = 6/7
What's the probability that B and -A? (1/2)(6/7) = 6/14
What's the probability that A and -B? (1/7)(1/2) = 1/14
Is 6/14 50% greater than 1/14? No, it's 500% greater.
D is wrong.

In B, the red was too small compared to the green.
In D, the red was too large compared to the green.
We need something in the middle.

Answer choice C.
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twobagels
Events A and B occur independently of one another. The probability that neither event occurs is 6 times the probability that both events occur. Furthermore, the probability that event B occurs and event A does not occur is 50 percent greater than the probability that event A occurs but event B does not occur. What is the probability that event B occurs?

A. 1/5
B. 1/4
C. 1/3
D. 1/2
E. 3/4

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Events A and B occur independently of one another. The probability that neither event occurs is 6 times the probability that both events occur. Furthermore, the probability that event B occurs and event A does not occur is 50 percent greater than the probability that event A occurs but event B does not occur.

What is the probability that event B occurs?

Let the probability of event A & event B occurring be a & b respectively.
The probability that neither event occurs = (1-a)(1-b)
The probability that both events occur = a*b

The probability that neither event occurs is 6 times the probability that both events occur.
(1-a)(1-b) = 6ab
1 - a - b + ab = 6ab
1 - a = 5ab + b = b(5a+1)
b = (1-a)/(5a+1)

The probability that event B occurs and event A does not occur = (1-a)b
The probability that event A occurs but event B does not occur = a(1-b)

The probability that event B occurs and event A does not occur is 50 percent greater than the probability that event A occurs but event B does not occur.
(1-a)b = 1.5a(1-b)
b - ab = 1.5a - 1.5ab
b + .5ab = 1.5a
b = 1.5a/(1+.5a)

b = (1-a)/(5a+1) = 1.5a/(1+.5a)
(1-a)(1+.5a) = 1 -a + .5a - .5a^2 = 1.5a(5a+1) = 7.5a^2 + 1.5a
1 - .5a - .5a^2 = 7.5a^2 + 1.5a
8a^2 + 2a - 1 = 0
8a^2 + 4a - 2a - 1 = 0
4a(2a + 1) - (2a + 1) = 0
(4a-1)(2a+1) = 0
a = 1/4 or - 1/2; Since negative probability is not possible
a = 1/4

b = (1-a)/(5a+1) = 1/3

IMO C
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