GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

It is currently 01 Apr 2020, 21:27

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

A fair coin is tossed 5 times. What is the probability of getting at

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Director
Director
User avatar
S
Joined: 17 Dec 2012
Posts: 619
Location: India
Re: A fair coin is tossed 5 times. What is the probability of getting at  [#permalink]

Show Tags

New post 15 May 2016, 20:51
gocoder wrote:
SravnaTestPrep wrote:
In this type of problems where repetition of values is allowed, the total number of possibilities is given by the formula \(n^r\) where n is 2 and has the values Heads and Tails. r is 5 and is equal to the number of tosses.

1. Total number of possibilities = \(2^5 = 32\)
2. Instances of favorable outcomes:

(i) HHH** - Let us elaborate all the possibilities:

1. HHH HH
2. HHH HT
3. HHH TH
4. HHH TT

We have exhausted the possibilities.

(ii) We also have the following **HHH and the possibilities are:

5. HH HHH
6.HT HHH
7,TH HHH
8.TT HHH

(iii) and the following: *HHH*

9. H HHH H
10, H HHH T
11. T HHH H
12. T HHH T

Out of the total 12 , only 8 are unique.

3. The probability is \(8/32 = 1/4.\)

How would you use this technique for atleast 2 heads ?


Hi,

In that case it is better to compute the opposite i.e., number of cases where there is only 1 head and the number of cases where there are two heads and three heads but not together and subtract it from the total number of cases.
_________________
Srinivasan Vaidyaraman
Magical Logicians
https://magical-logicians-sat-gmat-gre.business.site

Holistic and Holy Approach
Intern
Intern
avatar
B
Joined: 11 Aug 2013
Posts: 45
Re: A fair coin is tossed 5 times. What is the probability of getting at  [#permalink]

Show Tags

New post 15 May 2016, 21:29
Bunuel wrote:
nusmavrik wrote:
A fair coin is tossed 5 times. What is the probability of getting at least three heads on consecutive tosses?

A 2/16
B 1/4
C 7/24
D 5/16
E 15/32


At least 3 heads means 3, 4, or 5 heads.

3 consecutive heads
5 cases:
HHHTT
THHHT
TTHHH
HTHHH
HHHTH

\(P=5*(\frac{1}{2})^5=\frac{5}{32}\).

4 consecutive heads
2 cases:
HHHHT
THHHH

\(P=2*(\frac{1}{2})^5=\frac{2}{32}\).

5 consecutive heads
1 case:
HHHHH

\(P=(\frac{1}{2})^5=\frac{1}{32}\).

\(P=\frac{5}{32}+\frac{2}{32}+\frac{1}{32}=\frac{8}{32}=\frac{1}{4}\).

Answer: B.



Hi Bunnel,

My understanding of all the cases :-

consecutive 3 heads

3!/2! considering 3 H as one entity making total of 3 objects out of which 2 T are similar ,which makes it 3 .
Probability :- 3 (1/2)^5
consecutive 4 heads :-

2! , considering all H as one object making it therefore in total 2 objects
Probability :- 2(1/2)^5
Consecutive 5 Heads :-
1
Probablity :- 1(1/2)^5

Total Probablity :- 6(1/2)^5

Where am I going wrong?
Manager
Manager
avatar
Joined: 22 Feb 2015
Posts: 74
Re: A fair coin is tossed 5 times. What is the probability of getting at  [#permalink]

Show Tags

New post 15 May 2016, 22:22
1
ravisinghal wrote:
My understanding of all the cases :-

consecutive 3 heads

3!/2! considering 3 H as one entity making total of 3 objects out of which 2 T are similar ,which makes it 3 .

I also did it in a similar way, but after reading the solution, realized that I was 3 Heads consecutively does not mean 3 Heads and 2 Tails. Following would also qualify:

HTHHH
HHHTH
Director
Director
User avatar
S
Joined: 17 Dec 2012
Posts: 619
Location: India
Re: A fair coin is tossed 5 times. What is the probability of getting at  [#permalink]

Show Tags

New post 16 May 2016, 02:57
A general solution for constraints problems with line arrangement :

1. Represent the constraints at the left most. The first constraint is 3 consecutive heads
HHH_ _
2. Fill up the right most blanks
It can be TT and so HHHTT
3. Shift together what is constrained, one position at a time to the right and count the number of valid cases.
Here they are HHHTT, THHHT, TTHHH. So there are 3 cases
4. See if what is not constrained can be reordered.
In this case it is TT and cannot be reordered.
If it can be reordered, repeat step 3.
5. See if what is not constrained can change
If can change, make the change and go to step 2 else stop.
Here TT can change to TH or HT. Let us take TH.

Finally, we get the valid cases HHHTH and HTHHH. So there are 2 cases

The total number of cases for 3 consecutive Heads is 5.

Similarly we can do for the constraints 4 and 5 consecutive heads.
Note: When there are many constraints such as 3,4 or 5 consecutive heads start with the least constrained i.,e 3 here.
_________________
Srinivasan Vaidyaraman
Magical Logicians
https://magical-logicians-sat-gmat-gre.business.site

Holistic and Holy Approach
Intern
Intern
User avatar
B
Joined: 18 Apr 2017
Posts: 15
Location: Brazil
WE: Corporate Finance (Energy and Utilities)
Re: A fair coin is tossed 5 times. What is the probability of getting at  [#permalink]

Show Tags

New post 28 Dec 2017, 11:39
Bunuel - I did it using this approach - is it correct?

At least two heads = Exactly 3 consec. heads or Exactly 4 consec. heads or Exactly 5 consec. heads
At least two heads = P(3 consec. heads) + P(4 consec. heads) + P(5 consec. heads).

1) P(3 consec. heads):

1/2 * 1/2 * 1/2 = 1/8

2) P(4 consec. heads):

1/2 * 1/2 * 1/2 * 1/2 = 1/16

3) P(5 consec. heads):

1/2 * 1/2 * 1/2 * 1/2 * 1/2 = 1/32

That said:

1/8 + 1/16 + 1/32 = 1/4

In my point of view, in each scenario (3 consec heads for example), order does not matter:

HHHTT
THHHT
TTHHH

That is why I didn't bothered to count them using combinations.

Again - is this approach correct?
Thank you very much!
Manager
Manager
avatar
B
Joined: 23 Oct 2017
Posts: 62
Re: A fair coin is tossed 5 times. What is the probability of getting at  [#permalink]

Show Tags

New post 28 Dec 2017, 18:27
Total number of outcomes = 2^5 = 32
Favourable outcomes i.e. getting HHH
3 consecutive Hs =5
HHHTT, THHHT, TTHHH
HHHTH, HTHHH

4 consecutive Hs =2
HHHHT, THHHH

5 consecutive Hs = 1 (all)
HHHHH

probability = (1+2+5)/32 = 1/4
Director
Director
avatar
P
Joined: 02 Oct 2017
Posts: 686
Re: A fair coin is tossed 5 times. What is the probability of getting at  [#permalink]

Show Tags

New post 28 Apr 2018, 19:16
1
More than 3 consecutive heads

3 heads 2 tails
5!/(3!2!)=10 for consecutive cases are half so 5 cases

4 head 1 tail
5!/(4!1!)=5 ,for consecutive cases r half so 2 cases

5 head
1 case only

Total valid cases=8

Total cases=2*2*2*2*2 ( as each coin can take either H or T)

Result= 8/32=1/4

Posted from my mobile device
Intern
Intern
avatar
B
Joined: 17 Jun 2019
Posts: 33
Location: Israel
Concentration: Technology, Leadership
GPA: 3.95
WE: Engineering (Computer Hardware)
Reviews Badge
Re: A fair coin is tossed 5 times. What is the probability of getting at  [#permalink]

Show Tags

New post 14 Sep 2019, 03:58
another way to think about this is as blocks:
block of 5H is like 1C1 = 1
block of 4H is like 2C1 = 2
block of 3H is like 3C1 + 2 = 5
all possibilities is 2^5 = 32 ==> (1+2+5)/32=1/4
GMAT Club Bot
Re: A fair coin is tossed 5 times. What is the probability of getting at   [#permalink] 14 Sep 2019, 03:58

Go to page   Previous    1   2   [ 28 posts ] 

Display posts from previous: Sort by

A fair coin is tossed 5 times. What is the probability of getting at

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne