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.


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)


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
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sthahvi

Fantastic and hard Algebra question.


Let the:

Prob. A occurs = A

Prob. A NOT occurs = X

Prob. B occurs = B

Prob. B NOT occurs = Y

Events A and B occur independently of each other. If we want to find the Probability that two independent events occur together, we multiply the respective Probabilities.

“Prob that neither event occurs is equal to 6 times the probability that both events occur”

(Prob of NOT A) * (Prob of NOT B) = (6) * (Prob of A) * (Prob of B)

XY = (6)AB

“Prob that event B occurs AND event A does NOT occur ———is 50% greater than ———the Prob that event A occurs AND Prob that event B does NOT occur.”

Again, since the events of one event occurring and the other event not occurring are Indpendent Events we can multiply the individual probabilities to find the Prob that they occur together (“AND” Probability)

BX = (1.5)AY

Since each variable represents a positive probability value, we can divide the 1st equation above by this equation


XY = (6)AB

___________

BX = (1.5)AY


The X and A cancels out in the NUM/DEN

We are left with the equation:

Y/B = (6B) / (1.5Y)


1.5 divides evenly into 6 four times

Y/B = (4B) / Y

Cross multiply

Y^2 = (4) (B)^2

Again, since we are dealing with positive probabilities, when we take the even root (square root) we do not have to worry about the negative root

Take the square root of both sides

Y = 2B

Or in Ratio Form:

Y/B = 2/1

B ——- stands for he Prob that event B DOES occur

Y ———-stands for the Prob that event B does NOT occur

Together, B and Y are completely exhaustive of the possibilities. In other words:

Prob of B occurring + Prob of B NOT occurring = 1

Y/B = 2/1

Is he same as

Y : B = 2 : 1

Y (B not occurring) is twice as likely to occur as B is

Which means out of the complete sample space, the Prob that B will occur = 1/3

Or, using algebra


Y + B = 1

Y = 2B

2B + B = 1 ———-> B = 1/3

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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

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

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 - 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

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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See attached

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