HOT Competition 2 Sep/8AM: What is the probability of getting four tai

Question

What is the probability of getting four tails and two heads when an unfair coin is tossed 6 times?

(1) When the unfair coin is tossed thrice, the probability of getting two tails and one heads is 2/9.

(2) When the unfair coin is tossed twice, the probability of getting one tails and one heads is 4/9.

M36-103

Bunuel · Accepted Answer

Official Solution:

What is the probability of getting four tails and two heads when an unfair coin is tossed 6 times? 
 
Say the probability of tails is  p  and the probability of heads is  (1-p) . The question asks to find the value of  P(TTTTHH)=\frac{6!}{4!2!}*p^4*(1-p)^2  (notice that we multiply by  \frac{6!}{4!2!}  because four tails and two heads can occur in  \frac{6!}{4!2!}  ways: TTTTHH, TTTHTH, TTHTTH, THTTTH, HTTTTH, ... 
 
(1) When the unfair coin is tossed thrice, the probability of getting two tails and one heads is  \frac{2}{9} .
 
 P(TTH)=\frac{3!}{2!}*p^2*(1-p)=\frac{2}{9} . From this we can find the value of  p^2*(1-p) , which when squared gives the value of  p^4*(1-p)^2 , so we can find the value of  P(TTTTHH)=\frac{6!}{4!2!}*p^4*(1-p)^2 . Sufficient.
 
(2) When the unfair coin is tossed twice, the probability of getting one tails and one heads is  \frac{4}{9} . 
 
 P(TH)=2!*p*(1-p)=\frac{4}{9} . From this we can find that the value of  p  is  \frac{1}{3}  or  \frac{2}{3} . These two values will give two different values of  P(TTTTHH)=\frac{6!}{4!2!}*p^4*(1-p)^2 . Not sufficient.

Answer: A

rgaur19931 · Answer

The probability of getting 4 tails and 2 heads is x^4*y^2, where x = probability of getting tail, y = probability of getting heads.

Statement 1 : x^2*y = 4/9, we can get the value of x^4*y^2 = 16/81

Statement 2 : x*y = 2/9, We cannot use this as X can be 1/2 and y can be 4/9 or x can be 1/4 and y can be 8/9. Not sufficient.

Hence, the correct answer is a.

Kinshook · Answer

Asked: What is the probability of getting four tails and two heads when an unfair coin is tossed 6 times?

Let us assume that probability of head =p
Probability of tail =1-p
The probability of getting four tails and two heads when an unfair coin is tossed 6 times = 6C2×p^2(1-p)^4=15p^2(1-p)^4

(1) When the unfair coin is tossed thrice, the probability of getting two tails and one heads is 2/9.
The probability of getting two tails and one heads = 3C1×p×(1-p)^2= 3p(1-p)^2=2/9.
p=2/3; 1-p=1/3
The probability of getting four tails and two heads when an unfair coin is tossed 6 times = 6C2×p^2(1-p)^4=15p^2(1-p)^4=20/243
SUFFICIENT

(2) When the unfair coin is tossed twice, the probability of getting one tails and one heads is 4/9.
The probability of getting one tails and one heads = 2C1×p×(1-p)=2p(1-p)=4/9.
p=2/3; 1-p=1/3
The probability of getting four tails and two heads when an unfair coin is tossed 6 times = 6C2×p^2(1-p)^4=15p^2(1-p)^4=20/243
SUFFICIENT

IMO D

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EatMyDosa · Answer

Let  T = Tail  and  H = Head

Favourable cases  (4 tails, 2 heads) = Arranging 6 items (four are same and remaining two are same)

= Arranging 6 items (4 Ts, 2Hs)

= 6!/4!*2!

=  15 ways

Since the  coin in unfair , so probability  P(T)  and  P(H)    will not be   =  1/2

Let  P(T) = t  and  P(H) = h

So, probability of arranging 4Ts and 2Hs will be =  (t^4 * 2^h) * 15  ------- (1)

Rephrasing ,  what is the value of  t  and  h  ?

Statement 1 : Tells us the  P (2Ts, 1H) = 2/9

Favourable cases  = Arranging 3 items (2Ts and 1 H)

= 3!/2!

=  3 ways  (TTH, THT, HTT)

So,

P (2T, 1H) = (t^2*h) * 3 , or

(t^2*h) * 3 = 2/9 (given)

t^2*h = 2/27

This is  possible when   t= 1/3  and  h = 2/3  ( no other values for t and h are possible )

SUFFICIENT

Eliminate option B, C, D

Statement 2 : Tells us the  P (1Ts, 1H) = 4/9

Favourable cases = Arranging 2 items (1Ts and 1 H)

=  2 ways  (HT, TH)

So,

P (1T, 1H) = (t*h) * 2, or

(t*h) * 2 = 4/9 (given)

t*h = 2/9

Multiple cases are possible:

Case 1 : t = 1/3 , h = 2/3 ---> will give t^4*h^2 = 4/729 (from equation 1)
Case 2: t = 2/3, h = 1/3 ---> will give t^4*h^2 = 16/729

Since we get  conflicting values , eliminate D

Option A  (correct)

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shuvodip04 · Answer

Probability of getting 4T and 2H when tossed 6 times.
When the coin is tossed 6 times the arrangement of 4T and 2H (TTTTHH) is  \frac{6!}{4!2!} = 15
Required probability=  {P(T)}^4{P(H)}^2*15

Statement I:

{P(T)}^2*P(H)*3=\frac{2}{9} 
 =>{P(T)}^2*P(H)=\frac{2}{27} 
Required probability=  {P(T)}^4{P(H)}^2*15= {\frac{2}{27}}^2*15 
Sufficient.

Statement II:

P(T)*P(H)*2=\frac{4}{9} 
 =>P(T)*P(H)=\frac{2}{9}

We can't find the required probability from this statement.
Insufficient.

Ans:A

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anthien128 · Answer

IMO A is the correct answer.

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

We need to find the solution for P(four tails and two heads) = 6!/(4!2!)*x^4*(1-x)^2 = 15* x^4*(1-x)^2

(1) When the unfair coin is tossed thrice, the probability of getting two tails and one heads is 2/9.
We are given that P(two tails and one heads) = 3!/(2!1!)*x^2*(1-x)=2/9
x^2*(1-x)= 2/27

—> 15* x^4*(1-x)^2 = 15*(2/27)^2
—> (1) is SUFFICIENT

2) When the unfair coin is tossed twice, the probability of getting one tails and one heads is 4/9.
We are given that P(two tails and one heads) = 2!*x*(1-x)=4/9
x*(1-x)=2/9
9x-9x^2=2
9x^2-9x+2=0
(3x-2)(3x-1)=0
x=2/3 or 1/3

Case 1:
15* x^4*(1-x)^2 = 15*(2/3)^4*(1/3)^2

Case 2:
15* x^4*(1-x)^2 = 15*(1/3)^4*(2/3)^2

—> (2) is INSUFFICIENT

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HarshavardhanR · Answer

https://gmatclub.com/forum/download/file.php?id=86993 --- Harsha GMAT-Club-Forum-5uubtjis.png

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