The events A and B are independent, the probability that eve

The events A and B are independent, the probability that event A occurs is greater than 0, and the probability that event A occurs is twice the probability that event B occurs. The probability that at least one of events A and B occurs is 8 times the probability that both events A and B occur. What is the probability that event A occurs?

Let us say probability of A occuring is a.
Let us say probability of B occuring is b.

a = 2b

Probability (either A or B or both) = 8 times Probability (A and B)
a*(1-b) + b*(1-a) + ab = 8*ab

Substituting a=2b in the second equation:

2b*(1-b) + b*(1-2b) + 2b*b = 8*2b*b

3b-2b^2 = 16b^2
3b = 18b^2
b = 3/18 = 1/6

So, a = 2b = 1/3

I see its your first day, welcome to the club
and already solving 700 level probability, way to go *thumbs up*
it took me 3 mins !!

Thanks Chetan. Yes, I feel slightly less miserable in Maths than I do in Verbal:)

I was oh so close to solving. If anyone sees, can you explain why Probability (either A or B or both) =
a*(1-b) + b*(1-a) + ab? I get the 'ab' part, but why is it a*(1 - b) + b*(1-a)?
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Hi Bunuel,

Could you please explain the logic behind the answer key? I understand that we use x for P(b) and 2x for P(a). However, given that the problem states that "at least one of events A and B occurs is 8 times the probability that both events A and B occur..", how can the OA state that you can just use the formula P(a) + P(b) - P(a and b)? The "at least one" would encompass not only the probability of JUST a (and JUST b) but also the probability of both. Thus, it should be P(a) and P(b) - P(a and b) = 8[P(a and b)].

What am I missing here?

OA:

"Approach Strategically:

Let A and B be events. Let P(A) be the probability that event A occurs, let P(B) be the probability that event B occurs, let P(A or B) be the probability that at least one of the events A and B occurs, and let P(A and B) be the probability that both events A and B occur. The formula to calculate the probability that at least one event occurs is P(A or B) = P(A) + P(B) − P(A and B), (note that you cannot just add the two probabilities together). When the events A and B are independent, P(A and B) = P(A)P(B). So when the events A and B are independent, the formula for the probability that at least one of the events A and B occurs is P(A or B) = P(A) + P(B) − P(A)P(B).

Since the events A and B are independent, the probability that both of the events A and B occur is (2x)(x) = 2x2. Then the probability that at least one of the events A and B occurs is 2x + x − (2x)(x) = 3x − 2x2. Since the probability that at least one of the events A and B occurs is 8 times the probability that both of the events A and B occur, we have the equation 3x − 2x2 = 8(2x2). Let's solve this equation for x. We have 3x − 2x2 = 16x2, 3x = 18x2, x = 6x2. Since x is positive, we can divide both sides of the equation x = 6x2 by x. Then 1 = 6x. Dividing both sides of the equation 1 = 6x by 6, we have . Now we want the probability that event A occurs, which is 2x. The probability that event A occurs is . Answer Choice (D) is correct."

VeritasKarishma chetan2u Bunuel

What do we mean by at least one of events A and B occurs

P(A∪B) = means P(A) or P(B) but not P(A∩B) = P(A) + P(B) - P(A∩B)--> does this mean at least one of events A and B occurs ?

Because I was approaching to a wrong answer by doing P(A) + P(B) + P(A∩B) thinking that at least meant either P(A) or P(B) or both P(A)&P(B)

However doing P(A∩B) = P(A) + P(B) - P(A∩B) gave me the correct answer.
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At least one of events A and B occurs means P(A∪B).
A∪B means A or B INCLUDING A∩B

In the formula:
P(A∪B) = P(A) + P(B) - P(A∩B)

you must subtract "P(A∩B)" because when you add "P(A) + P(B)" you are adding "P(A∩B)" twice. It is easy to understand it if you draw a Venn diagram.

P(A∪B) means A occurs or B occurs or both occur.

P(A∪B) = P(A) + P(B) - P(A∩B)

Note that the minus sign above subtracts out P(A∩B) because P(A∩B) is counted two times. P(A∩B) is a part of P(A) and P(A∩B) is also a part of P(B). So you subtract it out to count it only once. After subtracting P(A∩B), P(A∩B) is still included in P(A∪B).

In other words,
P(A) = P(Only A) + P(A∩B)
and
P(B) = P(Only B) + P(A∩B)

So P(A∪B) = P(A) + P(B) - P(A∩B) = P(Only A) + P(A∩B) + P(Only B) + P(A∩B) - P(A∩B)
P(A∪B) = P(Only A) + P(A∩B) + P(Only B)

So P(A∪B) includes only event A occurs, only event B occurs and Both occur.

I prefer Plugging In The Answers PITA over algebra...usually just as fast and usually a lower risk of making a silly mistake.
I usually like trying B and D and tend to pick whichever of those looks easier to work with. In this case (lucky...?), that's D.

D:
A=1/3. B=1/6.
At least one of A and B = 1-notAandnotB = 1-(2/3)(5/6) = 1-(10/18) = 8/18
Both A and B = (1/3)(1/6) = 1/18
Is that what we wanted? Yep.

Answer choice D.

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