M31-26

Question

When an  unfair  coin is tossed twice, the probability of getting one tails and one heads is  \frac{1}{3} . What is the probability of getting two heads and two tails if the coin is tossed 4 times?

A.  \frac{2}{3} 
B.  \frac{1}{3} 
C.  \frac{1}{4} 
D.  \frac{1}{6} 
E.  \frac{1}{9}

manlog · Accepted Answer

Here is my approach:

1) The probability of getting one heads(H) and one tails(T) is P(H)*P(T)*2=1/3. Multiplier two is from HT and TH possibilities. Then P(H)*P(T)=1/6

2) The probability of getting two heads is P(H)^2 and two tails is P(T)^2, then the joint probability is  * *6= *6. Multiplier 6 is from number of different outcomes = 4C2.

Overall = 1/36*6=1/6

Bunuel · Answer

Official Solution:

When an  unfair  coin is tossed twice, the probability of getting one tails and one heads is  \frac{1}{3} . What is the probability of getting two heads and two tails if the coin is tossed 4 times?

A.  \frac{2}{3} 
B.  \frac{1}{3} 
C.  \frac{1}{4} 
D.  \frac{1}{6} 
E.  \frac{1}{9}

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

akashbolster · Answer

is there any other way of doing this question?

Bunuel · Answer

Check here: when-an-unfair-coin-is-tossed-twice-the-probability-of-getting-one-198795.html

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

umangshah · Answer

Hi Bunuel,

Can you explain how it is 2 times P(T)*P(H)? Is it because there is two possibilities of getting 1H & 1T (ie. HT & TH) or something else that I am missing.

Thanks

Bunuel · Answer

Exactly. We can get one tail and one head in two distinct sequences: either (head followed by tail) or (tail followed by head). The probability for each sequence is x(1 - x). Therefore, the overall probability of getting one tail and one head, is x(1 - x) + x(1 - x) = 2x(1 - x).

BottomJee · Answer

I think this is a  high-quality  question and I agree with explanation.

MBAToronto2024 · Answer

KarishmaB 
Hey! I used the following approach: 
P(HT) + P(TH) = 1/3 -> 2x = 1/3 -> x = 1/6 
HHTT -> we have to select two H out of 4 -> 4!/2!2! = 6

P(HHTT) = 1/6 * 1/6 * 6 = 1/6

Let me know if that's correct.

KarishmaB · Answer

Yes, it is.

But note here:" HHTT -> we have to select two H out of 4 -> 4!/2!2! = 6 "

We are not selecting 2 H out of 4. We are arranging HTHT ( = 1/6 * 1/6) into all other possible arrangements in 4!/2!*2! ways.

gmatbd · Answer

I did not quite understand the solution. I need help in understanding the mid part of solution where P (two heads and two tails) is mentioned. Why are we squaring the X and (1-x)? In case of P (one tail and one head) the X and (1-x) are not squared. Thanks

Bunuel · Answer

Let's break it down simply and clearly.

First, a quick recap of what x means:

x = probability of getting heads 
 1 - x = probability of getting tails

1. Why is P(one head and one tail) =  2 * x * (1 - x) ?

When tossing the coin   twice  , you get one head and one tail in two different ways:

First toss heads, second toss tails - probability =  x * (1 - x)  
 First toss tails, second toss heads - probability =  (1 - x) * x

So total =  x * (1 - x) + (1 - x) * x = 2 * x * (1 - x)

2. Why is P(two heads and two tails) =  6 * x^2 * (1 - x)^2 ?

Now we toss the coin   four   times and want exactly two heads and two tails.

We need:

Two of the tosses to be heads:  x * x = x^2   
 The other two tosses to be tails:  (1 - x) * (1 - x) = (1 - x)^2

And we must multiply by how many different ways we can arrange two H's and two T's in four tosses:

This is a combinatorics problem: number of ways =  \frac{4!}{2! * 2!} = 6

So total probability =  6 * x^2 * (1 - x)^2 = 6 * (x * (1 - x))^2

So why is it squared now?

Because we need 2 heads and 2 tails, so:

Probability of heads happening twice:  x * x = x^2  
 Probability of tails happening twice:  (1 - x) * (1 - x) = (1 - x)^2

When you multiply:  x^2 * (1 - x)^2 = (x * (1 - x))^2

That's why the expression is squared for the 4-toss scenario, but not for the 2-toss one.

M31-26

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