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

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

Am I correct to conclude that we can not find the probability of getting 3 tails on 4 tosses?

4C3 * x*x*x*(1-x) = 4/3 * x^2

We would have to know the individual probabilities of x and (1-x)?

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Re: When an unfair coin is tossed twice, the probability of getting one [#permalink]
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Bambi2021 wrote:
Am I correct to conclude that we can not find the probability of getting 3 tails on 4 tosses?

Yes, that's correct. The issue here is that the information is symmetric -- we learn that the probability of getting one head and one tail on two tosses is 1/3. If heads and tails were equally likely, that probability would instead be 1/2, and 1/3 is nowhere close to 1/2 in a probability setting, so it must be true that one of the two outcomes (heads or tails) is much more likely than the other. The problem is we don't know which. If you use the quadratic formula (which you don't need on the GMAT), the information in the stem tells us that on a single toss, we'll get one of the two results almost 80% of the time, and the other only about 20% of the time. If tails happens 80% of the time, it will be quite likely we get 3 tails on 4 tosses. But if tails only happens 20% of the time, it would be very rare to get 3 tails on 4 tosses. So your question has two possible answers. The question in this thread is only answerable because it asks for a symmetric result (2 heads, 2 tails) so we don't need to know which of the two outcomes is more likely.
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Re: When an unfair coin is tossed twice, the probability of getting one [#permalink]
Given: When an unfair coin is tossed twice, the probability of getting one tails and one heads is 1/3.
Asked: What is the probability of getting two heads and two tails if the coin is tossed 4 times?

Let probably of getting head be h and of getting tail be t
h = 1 - t

2C1 * h * t = 2ht = 1/3

Probability of getting two heads and two tails = 4C2 * h^2 t^2 = 6 * 1/6^2 = 1/6

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

Isn't the x*(1-x) the probability of getting one heads and one tails on two coin tosses? If not, then what is x*(1-x) in this scenario? How should I think about x*(1-x)?
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Re: When an unfair coin is tossed twice, the probability of getting one [#permalink]
Given: 2p(1-p)=1/3
p(1-p)=1/6

What is 4C2 p^2(1-p)^2=6/6^2=1/6

Geometric probability

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Re: When an unfair coin is tossed twice, the probability of getting one [#permalink]
Since there are two ways of getting a head and a tail in two tosses, namely:

TH
HT

and the probabilities of each way are identical, the probability of a HT or a TH is:

1/6

Two heads and two tails in 4 tosses contains via multiplication two instances of the above probability multiplied together. So a given pattern of two heads and two tails has a probability of:

1/36

Two heads and two tails can be achieved in multiple ways whose arrangements are:

4!/2!2! = 6 ways:

THHT
HTTH
HTHT
THTH
HHTTT
THTH

So the total probability of achieving two heads and two tails in 4 tosses is:

6*1/36 = 1/6

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