Last visit was: 19 Jul 2025, 11:36 It is currently 19 Jul 2025, 11:36
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 July 2025
Posts: 102,625
Own Kudos:
Given Kudos: 98,235
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,625
Kudos: 742,754
 [34]
2
Kudos
Add Kudos
32
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 July 2025
Posts: 102,625
Own Kudos:
742,754
 [8]
Given Kudos: 98,235
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,625
Kudos: 742,754
 [8]
3
Kudos
Add Kudos
5
Bookmarks
Bookmark this Post
General Discussion
avatar
rgaur19931
Joined: 30 Jun 2020
Last visit: 31 Jan 2022
Posts: 5
Own Kudos:
26
 [4]
Given Kudos: 37
Location: India
Posts: 5
Kudos: 26
 [4]
4
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
Kinshook
User avatar
Major Poster
Joined: 03 Jun 2019
Last visit: 19 Jul 2025
Posts: 5,703
Own Kudos:
5,238
 [3]
Given Kudos: 161
Location: India
GMAT 1: 690 Q50 V34
WE:Engineering (Transportation)
Products:
GMAT 1: 690 Q50 V34
Posts: 5,703
Kudos: 5,238
 [3]
1
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
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

Posted from my mobile device
User avatar
EatMyDosa
Joined: 06 Jan 2017
Last visit: 01 Dec 2022
Posts: 85
Own Kudos:
109
 [4]
Given Kudos: 283
Location: India
Concentration: General Management, Finance
GPA: 3.33
Products:
Posts: 85
Kudos: 109
 [4]
3
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
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)

Posted from my mobile device
User avatar
shuvodip04
Joined: 05 Oct 2017
Last visit: 03 Mar 2022
Posts: 87
Own Kudos:
147
 [1]
Given Kudos: 103
Location: India
Concentration: Finance, International Business
GMAT 1: 660 Q48 V33
GMAT 2: 700 Q49 V35 (Online)
GPA: 4
WE:Analyst (Energy)
GMAT 2: 700 Q49 V35 (Online)
Posts: 87
Kudos: 147
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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

Posted from my mobile device
User avatar
anthien128
Joined: 09 Jan 2011
Last visit: 19 Jun 2021
Posts: 72
Own Kudos:
104
 [1]
Given Kudos: 58
Location: Viet Nam
Concentration: Technology, Entrepreneurship
Posts: 72
Kudos: 104
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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

Posted from my mobile device
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 37,447
Own Kudos:
Posts: 37,447
Kudos: 1,013
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Moderators:
Math Expert
102625 posts
455 posts