It is currently 22 Nov 2017, 02:47

### 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 June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Manager
Joined: 04 Mar 2012
Posts: 50

Kudos [?]: 298 [0], given: 10

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

### Show Tags

30 Apr 2012, 01:05
16
This post was
BOOKMARKED
00:00

Difficulty:

45% (medium)

Question Stats:

61% (01:05) correct 39% (01:02) wrong based on 547 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
[Reveal] Spoiler: OA

Kudos [?]: 298 [0], given: 10

Math Expert
Joined: 02 Sep 2009
Posts: 42302

Kudos [?]: 133013 [0], given: 12402

Re: A fair coin is tossed 4 times. What is the probability of [#permalink]

### Show Tags

30 Apr 2012, 01:17
Expert's post
16
This post was
BOOKMARKED
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.
_________________

Kudos [?]: 133013 [0], given: 12402

Manager
Joined: 30 May 2013
Posts: 185

Kudos [?]: 87 [1], given: 72

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, 19:41
1
KUDOS
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

Kudos [?]: 87 [1], given: 72

Manager
Joined: 09 Jun 2013
Posts: 53

Kudos [?]: 70 [0], given: 3

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, 23:31
2
This post was
BOOKMARKED
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

Kudos [?]: 70 [0], given: 3

Manager
Joined: 10 Jun 2015
Posts: 126

Kudos [?]: 31 [0], given: 0

Re: A fair coin is tossed 4 times. What is the probability of [#permalink]

### Show Tags

14 Aug 2015, 00:39
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

Kudos [?]: 31 [0], given: 0

Manager
Joined: 13 Dec 2013
Posts: 172

Kudos [?]: 28 [0], given: 122

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

08 Dec 2016, 00:47
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

Kudos [?]: 28 [0], given: 122

Intern
Joined: 29 Nov 2016
Posts: 10

Kudos [?]: [0], given: 141

Re: A fair coin is tossed 4 times. What is the probability of [#permalink]

### Show Tags

16 Apr 2017, 00:40
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.

Kudos [?]: [0], given: 141

Manager
Joined: 01 Dec 2016
Posts: 118

Kudos [?]: 13 [0], given: 32

Location: Cote d'Ivoire
Concentration: Finance, Entrepreneurship
WE: Investment Banking (Investment Banking)
Re: A fair coin is tossed 4 times. What is the probability of [#permalink]

### Show Tags

22 May 2017, 09:10
This is a standard question, which should be approach with a general concept.
Otherwise, it is a waste of time to reconstruct the logic everytime we face this type of question.

The oncept is binomial variable probabilities.

Proba of having k success out-of n events for a binomial variable is given by a 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

Kudos [?]: 13 [0], given: 32

Re: A fair coin is tossed 4 times. What is the probability of   [#permalink] 22 May 2017, 09:10
Display posts from previous: Sort by