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# Two coins are tossed with probability landing at Head not

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22 Nov 2012, 19:09
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Question Stats:

53% (01:02) correct 47% (01:13) wrong based on 193 sessions

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Two coins are tossed with probability landing at Head not equal to 0.5. What is the probability of getting two Heads out of two tosses?

(1) The probability of getting head is 3 times that of getting tails
(2) Tossed two times, the probability of getting one Head and one Tail is 2/9
[Reveal] Spoiler: OA

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Last edited by Bunuel on 23 Nov 2012, 03:39, edited 1 time in total.
Renamed the topic and edited the tags.

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Re: Two coins are tossed with probability landing at Head not [#permalink]

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24 Nov 2012, 12:19
I got it right but i'm not sure if my guess was right.
As i looked at it, in (1) we can know what the odds of getting heads/tails are.
in (2) we could not know what the odds are for each one, only for both combined,
that's why (2) doesn't help at all....
did I get it right?

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Re: Two coins are tossed with probability landing at Head not [#permalink]

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26 Nov 2012, 07:35
ronr34 wrote:
I got it right but i'm not sure if my guess was right.
As i looked at it, in (1) we can know what the odds of getting heads/tails are.
in (2) we could not know what the odds are for each one, only for both combined,
that's why (2) doesn't help at all....
did I get it right?

Yes, that is correct!
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Aeros
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Re: Two coins are tossed with probability landing at Head not [#permalink]

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25 May 2014, 11:11
(1) Suppose we have a coin with 3 Head faces and 1 Tail face, a special biased coin.

Therefore probability of getting head in two tosses = 3/4*3/4= 9/16.

Therefore 1 is sufficient.
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Re: Two coins are tossed with probability landing at Head not [#permalink]

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31 May 2014, 13:03
If H+T = 1

Would the second statement be

2 (H / H+T) (T/H+T) = 2/9?

Then replacing we get that HT = 9
But still can't do the breakdown

Is this correct?

Cheers
J

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Re: Two coins are tossed with probability landing at Head not [#permalink]

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31 May 2014, 13:22
jlgdr wrote:
If H+T = 1

Would the second statement be

2 (H / H+T) (T/H+T) = 2/9?

Then replacing we get that HT = 9
But still can't do the breakdown

Is this correct?

Cheers
J

HT=(1/9) can be further simplified; as T=(9/h);

substituting the value of T in H+T=1 we will have a quadratic equation in H, which gives two different values of H , hence not sufficient.

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Re: Two coins are tossed with probability landing at Head not [#permalink]

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02 Jun 2014, 02:05
I got the first part(statement 1). As they are 100% we can find out that x%+y%=100%, and x%/y%=3 Gives us 75%/25%=3.

The second part I just "panicked" and couldn't solve. Will try and read up on probability.

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Re: Two coins are tossed with probability landing at Head not [#permalink]

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04 Jun 2014, 05:39
aeros232 wrote:
ronr34 wrote:
I got it right but i'm not sure if my guess was right.
As i looked at it, in (1) we can know what the odds of getting heads/tails are.
in (2) we could not know what the odds are for each one, only for both combined,
that's why (2) doesn't help at all....
did I get it right?

Yes, that is correct!

Hi,

Could you please explain the second statement a bit more clearly ? I mean why 2 is not sufficient?

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Re: Two coins are tossed with probability landing at Head not [#permalink]

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04 Jun 2014, 09:11
Expert's post
1
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BOOKMARKED
aniteshgmat1101 wrote:
aeros232 wrote:
ronr34 wrote:
I got it right but i'm not sure if my guess was right.
As i looked at it, in (1) we can know what the odds of getting heads/tails are.
in (2) we could not know what the odds are for each one, only for both combined,
that's why (2) doesn't help at all....
did I get it right?

Yes, that is correct!

Hi,

Could you please explain the second statement a bit more clearly ? I mean why 2 is not sufficient?

Two coins are tossed with probability landing at Head not equal to 0.5. What is the probability of getting two Heads out of two tosses?

Say the probability of heads is p and the probability of tails is 1-p.

(1) The probability of getting head is 3 times that of getting tails --> p=3(1-p) --> p=3/4 --> the probability of getting two Heads out of two tosses = 3/4*3/4. Sufficient.

(2) Tossed two times, the probability of getting one Head and one Tail is 2/9 --> P(HT)=2*p(1-p)=2/9 (we multiply by to because one Head and one Tail can occur in two ways HT and TH). We get two value for p. Not sufficient.

P.S. This is a poor quality question, since the values of p differ from (1) and (2), while on the GMAT, two data sufficiency statements always provide TRUE information and these statements never contradict each other or the stem.
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Re: Two coins are tossed with probability landing at Head not   [#permalink] 04 Jun 2014, 09:11
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