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Independent Events

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Independent Events

by manik11 » Thu Nov 19, 2015 4:18 am
Events A and B are independent and have equal probabilities of occurring. What is the probability that event A occurs?

1) The probability that at least one of events A and B occurs is .84.
2) The probability that event B occurs and event A does not is .24.

OA : A
Source : Veritas Prep

Even after going through the official explanation I can't understand why statement 2 is insufficient.
I also often get confused when the phrase "at least" is used in the probability questions.
Experts...could you guys please help me with this one?
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by GMATGuruNY » Thu Nov 19, 2015 4:51 am
manik11 wrote:Events A and B are independent and have equal probabilities of occurring. What is the probability that event A occurs?

1) The probability that at least one of events A and B occurs is .84.
2) The probability that event B occurs and event A does not is .24.
A and B have equal probabilities of occurring.
Thus:
A = B.
not A = not B.

P(at least A or B) + P(neither A nor B) = 1.
Put another way:
P(neither A nor B) = 1 - P(at least A or B).

Statement 1: The probability that at least one of events A and B occurs is .84.
Thus:
P(neither A nor B occurs) = 1 - 0.84 = 0.16.
(not A)(not B) = 0.16.

Since not B = not A, we get:
(not A)(not A) = 0.16
not A = 0.4.

Thus, A = 1 - 0.4 = 0.6.
SUFFICIENT.

Statement 2: The probability that event B occurs and event A does not is .24.
In other words:
(B)(not A) = 0.24.
Since B = A, we get:
(A)(not A) = 0.24.

This equation is satisfied by the values yielded in Statement 1.
If A = 0.6 and not A = 0.4, then (A)(not A) = (0.6)(0.4) = 0.24.
But this equation is also satisfied if A = 0.4 and not A = 0.6:
(A)(not A) = (0.4)(0.6) = 0.24.

Since A = 0.6 in the first case but A = 0.4 in the second case, INSUFFICIENT.

The correct answer is A.
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by [email protected] » Thu Nov 19, 2015 9:50 am
Hi manik11,

We're told that events A and B are independent and have EQUAL PROBABILITY of occurring. We're asked for the probability that event A occurs.

Fact 1: The probability that AT LEAST one of the events occurs is .84

This means that A, or B or BOTH occurs = .84

We can restate this to mean that the probability that neither A, nor B nor BOTH occurs = .16

(Not A)(Not B) = .16

Since we're told that A and B have equal probabilities of occurring, they also have equal probabilities of NOT occurring. So....

(Not A) = (Not B)

Combining these 2 equations gives us the probability that A will not occur.....

(.4)(.4) = .16

The probability that A DOES NOT occur = .4
The probability that A DOES occur = .6
Fact 1 is SUFFICIENT

Fact 2: The probability that B occurs and A DOES NOT occur = .24

This tells us that....

(B)(Not A) = .24

From the prompt, we know that...

A = B
(Not A) = (Not B)
A + (Not A) = 1

In this scenario, the values are .4 and .6, but we don't know which is which (one is the probability the event occurs, the other is the probability that the event does NOT occur). Either one COULD be the answer.
Fact 2 is INSUFFICIENT.

Final Answer: A

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by Max@Math Revolution » Sun Nov 22, 2015 1:10 am
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Events A and B are independent and have equal probabilities of occurring. What is the probability that event A occurs?

1) The probability that at least one of events A and B occurs is .84.
2) The probability that event B occurs and event A does not is .24.

There are 3 variables ( P(A), P(B), P(A&B)), and 2 equations (P(A)=P(B), P(A and B)=P(A)(B)), and 2 more equations are given from the 2 conditions, so there is high chance (D) will be the answer.
For condition 1, P(A or B)=P(A)+P(B)+P(A and B)=2P(A)-P(A)^2=0.84, P(A)^2-2P(A)+0.84=0. (P(A)-0.6)(P(A)-1.4)=0, and P(A)=0.6, 1.4, but probability cannot be greater than 1, so P(A)=0.6, and this is sufficient.
For condition 2, inP( only B)=P(B)-P(A and B), but we cannot know the value of P(B), so this is insufficient, and the answer becomes (A).

For cases where we need 1 more equation, such as original conditions with "1 variable", or "2 variables and 1 equation", or "3 variables and 2 equations", we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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