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

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Math Expert
Joined: 02 Sep 2009
Posts: 51215
When an unfair coin is tossed twice, the probability of getting one  [#permalink]

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26 May 2015, 09:29
1
23
00:00

Difficulty:

95% (hard)

Question Stats:

34% (02:09) correct 66% (01:42) wrong based on 172 sessions

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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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Math Expert
Joined: 02 Sep 2009
Posts: 51215
Re: When an unfair coin is tossed twice, the probability of getting one  [#permalink]

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01 Jun 2015, 02:09
1
5
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.

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Joined: 21 Feb 2012
Posts: 57
Re: When an unfair coin is tossed twice, the probability of getting one  [#permalink]

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27 May 2015, 06:41
3
3
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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J

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##### General Discussion
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Posts: 1325
Re: When an unfair coin is tossed twice, the probability of getting one  [#permalink]

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26 May 2015, 10:47
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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Math Expert
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Posts: 51215
Re: When an unfair coin is tossed twice, the probability of getting one  [#permalink]

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27 May 2015, 03:51
IanStewart wrote:
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.

Edited the options. Thank you.
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Re: When an unfair coin is tossed twice, the probability of getting one  [#permalink]

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23 Sep 2018, 23:44
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.

hi buneul

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

thanks
Math Expert
Joined: 02 Sep 2009
Posts: 51215
Re: When an unfair coin is tossed twice, the probability of getting one  [#permalink]

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24 Sep 2018, 00:10
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.

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.
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Re: When an unfair coin is tossed twice, the probability of getting one &nbs [#permalink] 24 Sep 2018, 00:10
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