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

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Director
Joined: 17 Dec 2012
Posts: 619
Location: India
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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.
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Srinivasan Vaidyaraman
Magical Logicians

Holistic and Holy Approach
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Re: A fair coin is tossed 5 times. What is the probability of getting at  [#permalink]

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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

5 cases:
HHHTT
THHHT
TTHHH
HTHHH
HHHTH

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

2 cases:
HHHHT
THHHH

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

1 case:
HHHHH

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

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

Hi Bunnel,

My understanding of all the cases :-

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

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

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

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

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

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
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Posts: 619
Location: India
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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.
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Srinivasan Vaidyaraman
Magical Logicians

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Re: A fair coin is tossed 5 times. What is the probability of getting at  [#permalink]

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

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

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

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
Joined: 23 Oct 2017
Posts: 62
Re: A fair coin is tossed 5 times. What is the probability of getting at  [#permalink]

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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
Joined: 02 Oct 2017
Posts: 686
Re: A fair coin is tossed 5 times. What is the probability of getting at  [#permalink]

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28 Apr 2018, 19:16
1

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

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

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

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Intern
Joined: 17 Jun 2019
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GPA: 3.95
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Re: A fair coin is tossed 5 times. What is the probability of getting at  [#permalink]

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14 Sep 2019, 03:58
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
Re: A fair coin is tossed 5 times. What is the probability of getting at   [#permalink] 14 Sep 2019, 03:58

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