GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 20 Jan 2019, 13:28

### 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

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

## Events & Promotions

###### Events & Promotions in January
PrevNext
SuMoTuWeThFrSa
303112345
6789101112
13141516171819
20212223242526
272829303112
Open Detailed Calendar
• ### FREE Quant Workshop by e-GMAT!

January 20, 2019

January 20, 2019

07:00 AM PST

07:00 AM PST

Get personalized insights on how to achieve your Target Quant Score.
• ### GMAT Club Tests are Free & Open for Martin Luther King Jr.'s Birthday!

January 21, 2019

January 21, 2019

10:00 PM PST

11:00 PM PST

Mark your calendars - All GMAT Club Tests are free and open January 21st for celebrate Martin Luther King Jr.'s Birthday.

# A fair coin is tossed 4 times. What is the probability of

Author Message
TAGS:

### Hide Tags

Intern
Joined: 04 Mar 2012
Posts: 37
A fair coin is tossed 4 times. What is the probability of  [#permalink]

### Show Tags

30 Apr 2012, 00:05
1
26
00:00

Difficulty:

45% (medium)

Question Stats:

64% (01:38) correct 36% (01:35) wrong based on 676 sessions

### HideShow timer Statistics

A fair coin is tossed 4 times. What is the probability of getting at least 2 tails?

A. 1/16
B. 1/2
C. 3/16
D. 11/16
E. 3/8
Math Expert
Joined: 02 Sep 2009
Posts: 52296
Re: A fair coin is tossed 4 times. What is the probability of  [#permalink]

### Show Tags

30 Apr 2012, 00:17
8
13
gmihir wrote:
A fair coin is tossed 4 times. What is the probability of getting at least 2 tails?

A. 1/16
B. 1/2
C. 3/16
D. 11/16
E. 3/8

Let's find the probability of the opposite event and subtract this value from 1.

The opposite event would be getting zero tails (so all heads) or 1 tail.

$$P(HHHH)=(\frac{1}{2})^4=\frac{1}{16}$$.

$$P(THHH)=\frac{4!}{3!}*(\frac{1}{2})^4=\frac{4}{16}$$, we are multiplying by $$\frac{4!}{3!}$$ since THHH scenario can occur in number of ways: THHH, HTHH , HHTH, or HHHT (notice that $$\frac{4!}{3!}$$ basically gives number of arrangements of 4 letters THHH out of which 3 H's are identcal).

$$P(T\geq{2})=1-(\frac{1}{16}+\frac{4}{16})=\frac{11}{16}$$.

Hope it's clear.
_________________
##### General Discussion
Manager
Joined: 30 May 2013
Posts: 155
Location: India
Concentration: Entrepreneurship, General Management
GPA: 3.82
Re: A fair coin is tossed 4 times. What is the probability of  [#permalink]

### Show Tags

01 Sep 2013, 18:41
2
gmihir wrote:
A fair coin is tossed 4 times. What is the probability of getting at least 2 tails?

A. 1/16
B. 1/2
C. 3/16
D. 11/16
E. 3/8

Another method:

Let's find the probability of 2 tails , 3 tails and all 4 tails

P(TTTT)=((1/2)^4=1/16.

P(HTTT)=(4!/3!)*(1/2)^4=4/16, we are multiplying by 4C3 (notice that {4!/3!} basically gives number of arrangements of 4 letters HTTT out of which 3 T's are identcal).

P(TTHH) = 4C2*(1/2)^4 = 4!/2!*2!*(1/2)^4=6/16

Total Probablity = 1/16 + 4/16 + 6/16
=11/16
Intern
Joined: 09 Jun 2013
Posts: 49
GMAT 1: 680 Q49 V33
GMAT 2: 690 Q49 V34
GPA: 3.86
Re: A fair coin is tossed 4 times. What is the probability of  [#permalink]

### Show Tags

01 Sep 2013, 22:31
1
gmihir wrote:
A fair coin is tossed 4 times. What is the probability of getting at least 2 tails?

A. 1/16
B. 1/2
C. 3/16
D. 11/16
E. 3/8

Here's another approach that I used:

Total possibilities = 2^4= 16 since each toss leads to 2 outcomes.

Way of getting no tail in 4 tosses = 1 (H H H H)
Way of getting 1 tail in 4 tosses = 4 (T H H H) (H T H H) ( H H T H) (H H H T)
So ways of getting at least two tails in 4 tosses = 11 (16-5)
Therefore, the probability = 11/16 (D).
_________________

Don't be afraid to fail, but be afraid not to try

Manager
Joined: 10 Jun 2015
Posts: 118
Re: A fair coin is tossed 4 times. What is the probability of  [#permalink]

### Show Tags

13 Aug 2015, 23:39
gmihir wrote:
A fair coin is tossed 4 times. What is the probability of getting at least 2 tails?

A. 1/16
B. 1/2
C. 3/16
D. 11/16
E. 3/8

Total outcome is 2^4=16
Number of cases of( tails < or =2) = 5, HHHH, THHH, HTHH, HHTH, HHHT
Therefore, the required probability is 11/16
Manager
Joined: 13 Dec 2013
Posts: 154
Location: United States (NY)
GMAT 1: 710 Q46 V41
GMAT 2: 720 Q48 V40
GPA: 4
WE: Consulting (Consulting)
Re: A fair coin is tossed 4 times. What is the probability of  [#permalink]

### Show Tags

07 Dec 2016, 23:47
1
Probability of 2T and 2H = 4C2*1/16 = 6/16
Probability of 3T and 1H = 4C3*1/16 = 4/16
Probability of 4T and 0H = 4C4*1/16 = 1/16

Summing the probabilities for each scenario that fulfils our requirements = 11/16
Intern
Joined: 29 Nov 2016
Posts: 9
Re: A fair coin is tossed 4 times. What is the probability of  [#permalink]

### Show Tags

15 Apr 2017, 23:40
gmihir wrote:
A fair coin is tossed 4 times. What is the probability of getting at least 2 tails?

A. 1/16
B. 1/2
C. 3/16
D. 11/16
E. 3/8

This is how I solved it, please tell me if it is correct.

Let's list the possibilities:
HHTT
HTHT
THHT
TTHH
THTH
HTTH

TTTH
TTHT
THTT
HTTT

TTTT

Now let's look at the first one: HHTT. Probability of getting H=1/2, H=1/2, T=1/2, T=1/2. So probability of getting HHTT = 1/2*1/2*1/2*1/2 = 1/16.
Similarly, for all the rest, the probability is 1/16. There are 11 possibilities on the list so the answer is 11*1/16 = 11/16.
Current Student
Joined: 01 Dec 2016
Posts: 109
Concentration: Finance, Entrepreneurship
GMAT 1: 650 Q47 V34
WE: Investment Banking (Investment Banking)
A fair coin is tossed 4 times. What is the probability of  [#permalink]

### Show Tags

22 May 2017, 08:10
This is a classic. it should be approach with a general concept.
Otherwise, it is a waste of time to reconstruct the logic everytime we face the same type of question.

The concept is binomial probabilities.

Proba of having k success out-of n events for a binomial variable has a given formula.
Let say, proba of success = p and proba of failure = (1-p)

Formula is nCk * (p^k)*(1-p)^(n-k)
if p=1/2 then formula becomes nCk/(2^n)

In the current question, p=1/2 and (1-p)=1/2
n=4 and p=2 or 3 or 4

Answer is (4C2+ 4C3 + 4C4)/(2^4) = 11/16

Option - D
_________________

What was previously considered impossible is now obvious reality.
In the past, people used to open doors with their hands. Today, doors open "by magic" when people approach them

Intern
Joined: 15 Feb 2017
Posts: 16
Location: India
Concentration: General Management, Finance
Schools: WHU MBA"20 (WL)
GMAT 1: 740 Q49 V41
GPA: 3.49
WE: Engineering (Other)
Re: A fair coin is tossed 4 times. What is the probability of  [#permalink]

### Show Tags

04 Dec 2017, 03:12
gmihir wrote:
A fair coin is tossed 4 times. What is the probability of getting at least 2 tails?

A. 1/16
B. 1/2
C. 3/16
D. 11/16
E. 3/8

I eliminated the options like follows:

A coin is tossed 4 times,
Probability of getting "at least 2" can be getting 2 tails, 3 tails and 4 tails.

Thus, probability >0.5.

There is only one option with probability >0.5

Please correct me if I am wrong, and yes I agree this is very crude approach.

I am very weak in probability.
Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 4551
Location: United States (CA)
Re: A fair coin is tossed 4 times. What is the probability of  [#permalink]

### Show Tags

12 Jan 2018, 06:31
gmihir wrote:
A fair coin is tossed 4 times. What is the probability of getting at least 2 tails?

A. 1/16
B. 1/2
C. 3/16
D. 11/16
E. 3/8

P(at least 2 tails) = 1 - P(not getting at least 2 tails)

P(at least 2 tails) = 1 - [P(getting exactly 1 tail) + P(getting 0 tails)]

Let’s look at P(getting exactly 1 tail):

P(T - H - H - H) = (1/2)^4 = 1/16

Since T - H - H - H can be arranged in 4!/3! = 4 ways, then P(getting exactly 1 tail) = 4 x 1/16 = 4/16.

Now let’s look at P(getting 0 tails), i.e., P(getting all 4 heads):

P(H - H - H - H) = (1/2)^4 = 1/16

P(at least 2 tails) = 1 - (4/16 + 1/16) = 11/16.

_________________

Scott Woodbury-Stewart
Founder and CEO

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Manager
Joined: 08 Sep 2016
Posts: 115
A fair coin is tossed 4 times. What is the probability of  [#permalink]

### Show Tags

12 Jan 2018, 14:22
1
Total outcome = 2^4 = 16

3 cases for favorable outcomes: (HHTT), (HTTT), (TTTT)
4C2 + 4C3 + 4C4 => 6 + 4 + 1= 11

Probability of at least 2 = (favorable outcome/ total outcome) = 11/16
A fair coin is tossed 4 times. What is the probability of &nbs [#permalink] 12 Jan 2018, 14:22
Display posts from previous: Sort by