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When an unfair coin is tossed twice, the probability of getting one

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When an unfair coin is tossed twice, the probability of getting one  [#permalink]

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New post 26 May 2015, 10:29
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When an unfair coin is tossed twice, the probability of getting one tails and one heads is 1/3. What is the probability of getting two heads and two tails if the coin is tossed 4 times?

A. 2/3
B. 1/3
C. 1/4
D. 1/6
E. 1/9

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Re: When an unfair coin is tossed twice, the probability of getting one  [#permalink]

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New post 01 Jun 2015, 03:09
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Bunuel wrote:
When an unfair coin is tossed twice, the probability of getting one tails and one heads is 1/3. What is the probability of getting two heads and two tails if the coin is tossed 4 times?

A. 2/3
B. 1/3
C. 1/4
D. 1/6
E. 1/9


OFFICIAL SOLUTION:

Let the probability of heads be x and the probability of tails be 1-x.

So, we are given that P(one tails and one heads) = 2*x*(1-x) = 1/3.

P(two heads and two tails) = 4!/(2!2!)*x^2*(1-x)^2 = 6*(x*(1-x))^2. We are multiplying by 4!/(2!2!) because HHTT scenario can occur in several ways: HHTT, HTHT, THHT, ... (permutation of 4 letters HHTT where 2 T's and 2 H's are the same).

Since from 2*x*(1-x) = 1/3 it follows that x*(1-x) = 1/6, then 6*(x*(1-x))^2 = 6*(1/6)^2 = 1/6.

Answer: D.
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Re: When an unfair coin is tossed twice, the probability of getting one  [#permalink]

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New post 27 May 2015, 07:41
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Hello all

My attempt:

Let \(x\) be the probability of getting a head. Therefore the probability of getting a tail will be \((1-x)\).
As per the given statement in the two tosses (successive events)
P(HT)+P(TH) = 1/3
\(x*(1-x) + (1-x)*x = 1/3\)
\(x*(1-x) = 1/6\)

For the new scenario we have the following combinations possible
\(P(4 tosses) = P(HHTT)+P(HTTH).....\)
In this case there will \(4!/(2!*2!)\) ways which is \(6\) ways.
Therefore
\(P(4 tosses) = 6*x^2*(1-x)^2\)
\(P(4 tosses) = 6*(1/6)^2\)
\(P(4 tosses) = 1/6\)

I will go with answer D.
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Re: When an unfair coin is tossed twice, the probability of getting one  [#permalink]

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New post 26 May 2015, 11:47
Bunuel, can you check the answer choices here? I don't think the right answer is among the choices, whether I interpret order to matter or not.
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Re: When an unfair coin is tossed twice, the probability of getting one  [#permalink]

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New post 27 May 2015, 04:51
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Re: When an unfair coin is tossed twice, the probability of getting one  [#permalink]

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New post 24 Sep 2018, 00:44
Bunuel wrote:
Bunuel wrote:
When an unfair coin is tossed twice, the probability of getting one tails and one heads is 1/3. What is the probability of getting two heads and two tails if the coin is tossed 4 times?

A. 2/3
B. 1/3
C. 1/4
D. 1/6
E. 1/9


OFFICIAL SOLUTION:

Let the probability of heads be x and the probability of tails be 1-x.

So, we are given that P(one tails and one heads) = 2*x*(1-x) = 1/3.

P(two heads and two tails) = 4!/(2!2!)*x^2*(1-x)^2 = 6*(x*(1-x))^2. We are multiplying by 4!/(2!2!) because HHTT scenario can occur in several ways: HHTT, HTHT, THHT, ... (permutation of 4 letters HHTT where 2 T's and 2 H's are the same).

Since from 2*x*(1-x) = 1/3 it follows that x*(1-x) = 1/6, then 6*(x*(1-x))^2 = 6*(1/6)^2 = 1/6.

Answer: D.



hi buneul


whats difference between fair and unfair coin??...

thanks
Math Expert
User avatar
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Joined: 02 Sep 2009
Posts: 50016
Re: When an unfair coin is tossed twice, the probability of getting one  [#permalink]

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New post 24 Sep 2018, 01:10
sdgmat89 wrote:
Bunuel wrote:
Bunuel wrote:
When an unfair coin is tossed twice, the probability of getting one tails and one heads is 1/3. What is the probability of getting two heads and two tails if the coin is tossed 4 times?

A. 2/3
B. 1/3
C. 1/4
D. 1/6
E. 1/9


OFFICIAL SOLUTION:

Let the probability of heads be x and the probability of tails be 1-x.

So, we are given that P(one tails and one heads) = 2*x*(1-x) = 1/3.

P(two heads and two tails) = 4!/(2!2!)*x^2*(1-x)^2 = 6*(x*(1-x))^2. We are multiplying by 4!/(2!2!) because HHTT scenario can occur in several ways: HHTT, HTHT, THHT, ... (permutation of 4 letters HHTT where 2 T's and 2 H's are the same).

Since from 2*x*(1-x) = 1/3 it follows that x*(1-x) = 1/6, then 6*(x*(1-x))^2 = 6*(1/6)^2 = 1/6.

Answer: D.



hi buneul


whats difference between fair and unfair coin??...

thanks


In case of fair coin P(heads) = P(tails) = 1/2. If a coin is not fair, then the probabilities might be different.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: When an unfair coin is tossed twice, the probability of getting one &nbs [#permalink] 24 Sep 2018, 01:10
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