probability

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probability

by jsasipriya » Thu Jun 24, 2010 4:31 pm
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?
Answer: 1/3
Can someone explain this?

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by Patrick_GMATFix » Thu Jun 24, 2010 6:00 pm
Hi Priya. I hope I can help

For any two independent events A and B, the following probability rules apply:

(1) Probability that both events occur is the product of the individual probabilities:
  • P(A and B) = P(A)*P(B)
(2) Probability that one or the other or both events occur is the sum of the probabililties of each minus the probability that both occur:
  • P(A or B) = P(A) + P(B) - [P(A)*P(B)]
With these concepts in mind, let's get to your question shall we? Below is the data we're given, and what it means

"the probability of A is twice that of B" --> P(A)=2x P(B)=x. Since the question ultimately asks for the probability of A, at the end of the day we want the value of 2x

"The probability that at least one occurs" --> P(A or B) = 2x + x - [2x*x]. We're told that this probability is "8 times the probability that both occur" --> = 8*[2x*x]

Putting this in an equation gives us: 2x + x - [2x*x] = 8*[2x*x].

Divide both sides by x to get --> 2 + 1 - 2x = 8*2x

Solve for x --> 3 = 18x --> x=1/6

We're asked to find the probability of A, or the value of 2x. 2x = 2*(1/6) --> 1/3

If you have trouble with probability questions, use the Drill Generator to create and take timed drills, and set topics='Combinatorics'

Hope that makes sense,
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by mj78ind » Thu Jun 24, 2010 8:28 pm
Say probabilities are a and b. Then per stem

1 - (1-a)*(1-b) = 8a*b

Note for independent events a and b occurring simultaneously is given by a*b. The 1 - (1-a)*(1-b) is used for alteast one of a or b to occur.

We also know a = 2b, substituting for b and solving for a and discarding the solution with a = 0 gives us only one answer 1/3.

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by sars72 » Thu Jun 24, 2010 9:36 pm
Patrick, awesome break down of the problem ...lucid explanation (as always)

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by Patrick_GMATFix » Fri Jun 25, 2010 6:53 am
sars72 wrote:Patrick, awesome break down of the problem ...lucid explanation (as always)
Careful SARS, you'll make me blush. Thank you for the kind words :-)

I see from your blog (congrats!!) that your test is in a week or so. Send me a PM if you would like free access to the GMATFix Solutions Engine. Depending on your study strategy, you may benefit from taking topic specific drills from GMATPrep questions.

By the way, why choose username 'sars'? you know that's the name of the Bird Flu right (stands for Severe Acute Respiratory Syndrome)?

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by GMATGuruNY » Fri Jun 25, 2010 8:46 am
jsasipriya wrote:
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?
Answer: 1/3
Can someone explain this?
We also could use the answer choices for this question, since we know that the correct probability has to be one of them.

First let's review some basic probabillity concepts:

P(A and B) = P(A) * P(B). When we want the probability of events happening together, we multiply the fractions, because the more things that we want to happen together, the smaller the probability. When we multiply fractions, the result just keeps getting smaller.

Also:

P(event happens) + P(event doesn't happen) = 1. If we have a 1/4 chance of choosing a red marble, we have a 1-(1/4) = 3/4 chance of not choosing a red marble.

So when a question asks for the probability that something happens at least once, remember:

So P(at least one) + P(none) = 1. Why? Because if the event doesn't happen -- if we don't get "at least one" -- then we're getting none.

Let's say we flip a coin twice, and we want the probability that we get heads at least once:

Since P(at least one heads) + P(no heads) = 1, then P(at least one heads) = 1 - P(no heads)

The probability of not getting heads on any given flip is 1/2. So for the two flips, P(no heads) = 1/2 * 1/2 = 1/4.

Thus, P(at least one heads) = 1 - 1/4 = 3/4.

Now onto the question above! Suppose we try the answer choice that says that P(A) = 1/3:

If P(A) = 1/3, then P(B) = 1/2 * 1/3 = 1/6. [Because the problem says that P(A) is twice P(B), which means P(B) = 1/2 * P(A).]

What's the probability that at least A or B occurs? Remember the rule above:

P (at least one) + P (none) = 1.

So P(at least A or B) + P(not A and not B) = 1, and P(at least A or B) = 1 - P(not A and not B).

If P(A) = 1/3 and P(B) = 1/6, then P(not A) = 2/3 and P(not B) = 5/6, so P(not A and not B) = 2/3 * 5/6 = 10/18.

Using the rule above, P(at least A or B) = 1 - 10/18 = 8/18.

Is this 8 times the probability of getting both A and B? Yes:

P(A and B) = 1/3 * 1/6 = 1/18. Since 8/18 = 8 * 1/18, we have found the correct answer. Success!

P(A) = 1/3.
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by boysangur » Tue Nov 16, 2010 9:38 pm
If it says "the probability of at least one event" doesn't it mean "probability of one happening or the other or both"? In which case, P (at least 1) = P(A) + P(B) + P(AB). I know it's wrong but could someone explain why?

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by diebeatsthegmat » Wed Nov 17, 2010 6:13 pm
Patrick_GMATFix wrote:Hi Priya. I hope I can help

For any two independent events A and B, the following probability rules apply:

(1) Probability that both events occur is the product of the individual probabilities:
  • P(A and B) = P(A)*P(B)
(2) Probability that one or the other or both events occur is the sum of the probabililties of each minus the probability that both occur:
  • P(A or B) = P(A) + P(B) - [P(A)*P(B)]
With these concepts in mind, let's get to your question shall we? Below is the data we're given, and what it means

"the probability of A is twice that of B" --> P(A)=2x P(B)=x. Since the question ultimately asks for the probability of A, at the end of the day we want the value of 2x

"The probability that at least one occurs" --> P(A or B) = 2x + x - [2x*x]. We're told that this probability is "8 times the probability that both occur" --> = 8*[2x*x]

Putting this in an equation gives us: 2x + x - [2x*x] = 8*[2x*x].

Divide both sides by x to get --> 2 + 1 - 2x = 8*2x

Solve for x --> 3 = 18x --> x=1/6

We're asked to find the probability of A, or the value of 2x. 2x = 2*(1/6) --> 1/3

If you have trouble with probability questions, use the Drill Generator to create and take timed drills, and set topics='Combinatorics'

Hope that makes sense,
-Patrick
so nice solution. thanks. i forgot that P (a &b)=Pa*Pb :(