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

Question

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

Bunuel · Accepted Answer

OFFICIAL SOLUTION:

Let's denote the probability of getting heads as  x  and the probability of tails as  1-x .
 
We are given that  P(one \ tails \ and \ one \ heads) = 2*x*(1-x) = \frac{1}{3} .
 
The probability of getting two heads and two tails is given by  P(two \ heads \ and \ two \ tails) = \frac{4!}{(2!2!)}*x^2*(1-x)^2 = 6*(x*(1-x))^2 . Here, we multiply by  \frac{4!}{(2!2!)}  to account for the different ways that two heads and two tails can occur: HHTT, HTHT, THHT, etc. (this is a permutation of 4 letters HHTT where 2 T's and 2 H's are identical).
 
From  2*x*(1-x) = \frac{1}{3} , it follows that  x*(1-x) = \frac{1}{6} . Thus,  6*(x*(1-x))^2 = 6*(\frac{1}{6})^2 = \frac{1}{6} .

Answer: D

Jackal · Answer

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.

sdgmat89 · Answer

hi buneul

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

thanks

Bunuel · Answer

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

Bambi2021 · Answer

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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IanStewart · Answer

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.

Kinshook · Answer

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

Fatchoco · Answer

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)?

Oppenheimer1945 · Answer

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

Posted from my mobile device

Regor60 · Answer

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

Posted from my mobile device

freebie · Answer

Hii Bunuel!
Please check if the following method is right:
so as given Probability of one heads and one tail is 1/3:
we can say that TH + HT = 1/3 
so taking one case at a time would result in TH = 1/6
Now that we have 4 coins tossed where in we need a tail and head( TH or HT) in pair of 2 
which can be occurred in 4!/2!2! = 6 ways 
= 6 * 1/6 * 1/6 =1/6

AVMachine · Answer

Something Different:
Given Eq.1 as HT + TH = 1/3
Now we want HHTT -> 4!/(2! 2!) = 6 Such Combination

Now Multiply Eq. 1 with Eq. 1;

(HT+TH) * (HT+TH) = 1/9
HTHT + THHT + HTTH + THTH = 1/9

Now 4 ----------- 1/9
Then 6 ----------- 1*6/(9*4) = 1/6

Ans. D

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

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