conditional probability

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conditional probability

by sana.noor » Sat Mar 02, 2013 9:23 am
Denette tosses two coins. What is the probability that she has tossed 2 heads, given that she has tossed at least 1 head?
a)1/3
b)1/4
c)4/3
d)3/4
e)1/2

a
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by Brent@GMATPrepNow » Sat Mar 02, 2013 9:44 am
sana.noor wrote:Denette tosses two coins. What is the probability that she has tossed 2 heads, given that she has tossed at least 1 head?
a)1/3
b)1/4
c)4/3
d)3/4
e)1/2

a
Here's a non-formulaic approach.

I'll assume that the probability of tossing heads = the probability of tossing tails = 0.5
Since each outcome is equally likely, let's draw a tree diagram to see all of the possible outcomes:
Image

So, there are 4 possible outcomes and each outcome is equally likely.
The 4 outcomes are HH, HT, TH and TT

Now it's given that there's at least one head.
So, of the 4 possible outcomes, only 3 of them are such that there's at least one head.
Image

Now that we know we've narrowed the field to three equally-likely outcomes (HH, HT and TH), we want the probability of HH
Image

The probability is [spoiler]1/3 (answer = A)[/spoiler]

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by sana.noor » Sat Mar 02, 2013 10:22 am
Brent should one practice Binomal theorem to find the probability? is it important to study for gmat or one can skip?
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by Brent@GMATPrepNow » Sat Mar 02, 2013 10:24 am
Well, you could use the binomial theorem for some questions, but the GMAT would never ask you a question that requires it.
I'd skip it.

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by GMATGuruNY » Sat Mar 02, 2013 1:54 pm
sana.noor wrote:Denette tosses two coins. What is the probability that she has tossed 2 heads, given that she has tossed at least 1 head?
a)1/3
b)1/4
c)4/3
d)3/4
e)1/2

a
Here are all of ways to get at least one heads:
HT
HH
TH

Of these 3 options, one the outcome in red is favorable, implying that the probability = 1/3.

The correct answer is A.
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by Anurag@Gurome » Sat Mar 02, 2013 7:28 pm
sana.noor wrote:Denette tosses two coins. What is the probability that she has tossed 2 heads, given that she has tossed at least 1 head?
Number of outcomes with at least one head = 3 {HH, TH, HT}

Number of outcomes with two heads = 1 {HH}

Hence, required probability = 1/3

The correct answer is A.
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