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.

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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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.

A

4T and 2H when unfair coin is tossed 6 times

1:
when coin is tossed thrice then we have probability of 2T and 1 H
If we need to calculate 6 times then we can just multiple
e.g.
TTH = we have
we need to find
TTH TTH = 2/9*2/9
hence sufficient


2:
if coin is tossed twice:
1h1T= 4/9
if we multiple for 6 times then we can only get 3H and 3T probability
so it is hard to find probability of 4T and 2 H when coin is tossed 6 times
insufficient

Hence A is our answer

A

I assumed x = p(H)
And 1-x = p(T)

X^2(1-X) = 2/9

We need to find x^4(1-x)^2
So simple square it.

2 is not sufficient

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Q. 6C4(1/p1)^4 * (1/p2)^2 = ?
p1 =prob (T)
p2 = prob(H)
p1+p2 = 1

St1
3C2(1/p1)^2(1/p2) = 2/9
(1/p1)^2(1/p2) = 2/27
Sufficient

St.2
2C1(1/p1)(1/p2) = 4/9
1/(p1*p2) = 2/9
1/p1(1-p1) = 2/9
No Solution
Insufficient

Hence A

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As, it is an unfair coin, we need to find the probability for this one, as normal 1/2 probability will not work here.

Statement A --> 2 T & 1 H - P=2/9. For normal coin, it will be 3/8.
As we need to find probability for 4T & 2H - it will be same as 2T & 1H - Hence, statement A is sufficient.

Statement B -- 1T & 1H P=4/9. Tells nothing, if coin is heavier oon which side, and hence NOT SUFFICIENT.

Option A

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Answer Option A
To find - P(H)²P(T)⁴

(1) When the unfair coin is tossed thrice, the probability of getting two tails and one heads is 2/9.
P(H)P(T)² = 2/9
(P(H)P(T)²)² = (2/9)²
P(H)²P(T)⁴ = 4/81

SUFFICIENT

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

P(H)P(L)= 4/9

Not sufficient to deduce P(H)²P(T)⁴

INSUFFICIENT

ANSWER - A

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I have no idea how to solve this, but I took a wild guess and chose A, since the number of tosses and combination of H and T are twice of that in option A.

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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

Solution:

Let us think that unfair coin means a coin that has either both heads or both tails.
So we are asked to find P(4T & 2H) when an unfair coin is tossed 6 times

Statement 1:

Given P(2T & 1H) = 2/9 when an unfair coin is tossed thrice.
From this data, we cannot deduce exactly P(4T & 2H). So statement 1 alone is not sufficient to answer the question.

Statement 2:

Given P(1T & 1H) = 4/9 when an unfair coin is tossed twice.
From this data, we cannot deduce exactly P(4T & 2H). So statement 2 alone is not sufficient to answer the question.

Even after combining statement 1 & statement 2, we cannot find the probability P(4T & 2H) when the coin is tossed 6 times.

So option E is the correct answer.

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we have to find the probability of 4 Tails and 2 Heads =T*T*T*T*H*H

ST1.2 tails and 1 head =2/9
T*T*H =2/9
by this we can calculate the prob of 4T and 2H ie 2/9*2/9
Sufficient

ST2 .1 t and 1 head =4/9

this is not sufficient

OA:a

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

we have to find: 6C4 * P(T)^4 *P(H)^2 = ?


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

Since, we know 3C2 * P(T) ^ 2 * P(H) = 2/9 => P(T)^2 * P(H)= 2/27. From this data we can find our answer to the required question.

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

Since, from statement (2) we get 2C1 * P(T) * P(H) = 4/9 => P(T) * P(H) = 2/9. From this information we cannot find the answer to our question.

Hence, A is the correct answer.

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.
2Tails*Head=T*T*H=2/9
we cant make any clear conclusion from this regarding individual probabilities.
so insufficient.

(2) When the unfair coin is tossed twice, the probability of getting one tails and one heads is 4/9.
T*H=4/9
We can infer that T=H=2/3
thus we can calculate four tails and two heads=\((2/3)^6\)

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

Let H be the probability of Head and T be the probability of Tails

(1) When the unfair coin is tossed thrice, the probability of getting two tails and one heads is 2/9.
- Notice that we are asked the probability of 4 tails + 2 heads out of 6 tosses; the outcome and number of tosses are exactly doubled.
from 1), we get: 3C1 * \(T^2 * H = \frac{2}{9}\)
\(T^2 * H = \frac{2}{27}\)

We are asked the following: 6C2* \(T^4 * H^2\)
that is \(15 T^4 H^2\)
Since we know \(T^2 * H = \frac{2}{27}\), \(15 T^4 H^2\)= \(4*15 / 27^2\)
- SUFFICIENT

(2) When the unfair coin is tossed twice, the probability of getting one tails and one heads is 4/9.
This will get us value of T*H, but we need \(T^4 * H^2\)
- INSUFFICIENT

Answer A.

Let probability of heads = h
=> probability of tails = (1-h)


P of getting 2 heads and 1 tail = h^2 (1-h)
This can happen in 3 ways, HHT HTH THH
So, total probability is 3 x h^2 (1-h)
Given: 3 x h^2 (1-h) = 2/9 .... (eqn 1)
This clearly has multiple solutions (because of h^3). So, not sufficient


Along the lines of statement 1,
2 x h x (1-h) = 4/9 .... (eqn 2)
Again, this is a quadratic eqn. Thus multiple possible solutions. So, not sufficient.

Now taking both together...
Dividing eqn 1 by eqn 2 we get:
(3 h^2 (1-h) )/ 2h(1-h) = (2/9) / (4/9)
=> 3h/2 = 1/2
=> h = 1/3

We can cancel out the h and (1-h) terms because we know they aren't 0. Else eqn 1 and 2 would fail.

So, the statements are together sufficient.
Answer: C

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Answer:A
Question: probability of getting 4 tails and 2 heads=6C4*x^4*(1-x)^2=?
Let probability of getting tail=x and probability of getting head=1-x
Statement-1: sufficient
Given probability of getting two tail and one head=2/9
3C2*x^2*(1-x)=2/9
x^2*(1-x)=2/27
Now Question: probability of getting 4 tails and 2 heads=6C4*x^4*(1-x)^2=15*((x^2*(1-x))^2=15*(2/27)^2=20/243
Statement-2:not sufficient
Given 2C1*x*(1-x)=4/9
9x^2-9x+2=0
x=2/3 and 1/3
Question: probability of getting 4 tails and 2 heads=6C4*x^4*(1-x)^2=80/81 for x=2/3
Question: probability of getting 4 tails and 2 heads=6C4*x^4*(1-x)^2=20/243 for x=1/3
Not sufficient

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let P of getting H x and that of tails 1-x
so 4 T and 2 H ; 6!/4!*2! ;
15 *(x)^2*(1-x)^4--(a)
#1
When the unfair coin is tossed thrice, the probability of getting two tails and one heads is 2/9.
TTH ; 3!/2! ; 3 *x^1*(1-x)^2 = 2/9 ;
x^1*(1-x)^2 = 2/27
(1-x)^2 = 2/27x
substitute in (a)
15 *(x)^2*(1-x)^4 ;
15*x^2*( 2/27x)^2 ;
15*4/(27^2) ;
sufficient

#2
When the unfair coin is tossed twice, the probability of getting one tails and one heads is 4/9
1 T 1 Head ; 2
2*x*(1-x) = 4/9 ;
x*(1-x) =2/9
1-x = 2/9x
not sufficient as value of x wont be matching with square
IMO A is correct




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.

Let P{Tails} be x/y
And P{Heads} be y-x/y

St 1)
2T and 1H can occur in three ways (TTH, THT, HTT)
{TTH} = x*x*(y-x)/y^3 -(i)
3{TTH} = 2/9
{TTH} = 2/27 -(ii)

From (i) and (ii)
x = 1 and y = 3

P{Tails} = 1/3
P{Heads} = 2/3

SUFFICIENT

St 2)
1T and 1H can occur in two ways (TH, HT)
{TTH} = x*(y-x)/y^2 -(iii)
2{TH} = 4/9
{TH} = 2/9 -(iv)

From (iii) and (iv)
x = 1 or 2 , y = 3

P{Tails} = 1/3 AND P{Heads} = 2/3
OR
P{Tails} = 2/3 AND P{Heads} = 1/3

INSUFFICIENT

Correct Answer is A

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We need to find probability of 4T and 2H in 6 tosses:

(1) When the unfair coin is tossed thrice, the probability of getting two tails and one heads is 2/9.
P(2T & 1H) = 2/9
We can repeat the same probability in the next 3 tosses to get the desired outcome.
So, P(4T & 2H) =P(2T & 1H)*P(2T & 1H)
= 2/9*2/9
= 4/81
Sufficient

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

In this case, we will not be able to fi d the required probability. We can only find probability of 3T and 3H.
Not sufficient

IMO A

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