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A fair coin is tossed 5 times.

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A fair coin is tossed 5 times. [#permalink] New post 07 Jul 2017, 12:06
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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
[Reveal] Spoiler: OA

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Last edited by pranab01 on 26 Jul 2017, 04:37, edited 3 times in total.
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Re: A fair coin is tossed 5 times. [#permalink] New post 26 Jul 2017, 00:24
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Any explanation on the above? Thanks
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Re: A fair coin is tossed 5 times. [#permalink] New post 26 Jul 2017, 04:49
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nancyjose wrote:
Any explanation on the above? Thanks


The total number of outcomes = 2 ^5 =32 (it is because each toss has two possibilities Head or Tail.In general when a coin is tossed n times , the total number of possible outcomes = 2^n)

Let E = event of getting atleast 3 heads .

Atleast means that’s the minimum number of heads and it can be more than that ..ie he can get 3 heads , 4 heads or 5 heads.

So Favorable outcomes E ={3 heads and remaining 2 tails , 4 heads and 1 tail , 5 heads no tail}
={HHHTT,TTHHH,THHHT,HHHHT,THHHH,HTHHH,HHHTH, HHHHH}

So no. of favorable outcomes E =8
So the required probability = 8/32 = ¼
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Re: A fair coin is tossed 5 times. [#permalink] New post 26 Jul 2017, 06:32
pranab01 wrote:
nancyjose wrote:
Any explanation on the above? Thanks


The total number of outcomes = 2 ^5 =32 (it is because each toss has two possibilities Head or Tail.In general when a coin is tossed n times , the total number of possible outcomes = 2^n)

Let E = event of getting atleast 3 heads .

Atleast means that’s the minimum number of heads and it can be more than that ..ie he can get 3 heads , 4 heads or 5 heads.

So Favorable outcomes E ={3 heads and remaining 2 tails , 4 heads and 1 tail , 5 heads no tail}
={HHHTT,TTHHH,THHHT,HHHHT,THHHH,HTHHH,HHHTH, HHHHH}

So no. of favorable outcomes E =8
So the required probability = 8/32 = ¼



Thank you this was helpful
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Re: A fair coin is tossed 5 times. [#permalink] New post 04 Jan 2019, 23:26
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Expert's post
nancyjose wrote:
pranab01 wrote:
nancyjose wrote:
Any explanation on the above? Thanks


The total number of outcomes = 2 ^5 =32 (it is because each toss has two possibilities Head or Tail.In general when a coin is tossed n times , the total number of possible outcomes = 2^n)

Let E = event of getting atleast 3 heads .

Atleast means that’s the minimum number of heads and it can be more than that ..ie he can get 3 heads , 4 heads or 5 heads.

So Favorable outcomes E ={3 heads and remaining 2 tails , 4 heads and 1 tail , 5 heads no tail}
={HHHTT,TTHHH,THHHT,HHHHT,THHHH,HTHHH,HHHTH, HHHHH}

So no. of favorable outcomes E =8
So the required probability = 8/32 = ¼



Thank you this was helpful


Good solution

I have a little trick that helps prevent getting lost in repetition when listing HHHTT, THHHT.. and what not..
I just replace 3 heads in a row with A.
then my list of outcomes looks like this
ATT
TAT
TTA
HTA
ATH
THA
AHT
AHH

I hope this helps save some time and headache
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Re: A fair coin is tossed 5 times. [#permalink] New post 06 Jan 2019, 00:14
HappyMathtutor wrote:
Good solution

I have a little trick that helps prevent getting lost in repetition when listing HHHTT, THHHT.. and what not..
I just replace 3 heads in a row with A.
then my list of outcomes looks like this
ATT
TAT
TTA
HTA
ATH
THA
AHT
AHH

I hope this helps save some time and headache


Awesome trick, really helpful
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Re: A fair coin is tossed 5 times. [#permalink] New post 06 Jan 2019, 08:40
Expert's post
pranab01 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



Ofcourse, we could use a mix of combinations and logic too..

1) all 5 heads - 1 HHHHH
2) 4 heads - ways 5C4=5... Only one where T is in centre should be discarded HHTHH so 5-1=4
2) 3 heads - let 3 heads be one H so HTT, THT and TTH - so 3 ways
total - 1+4+3=8

Also the trick of Happymathtutor is good.
I would use this in combinations too solve..
Take all H to be 1 as h
1) all 5 heads - hHH - 1 way
2) 4 heads hHT - so ways = 3*2*1=6 but hH and Hh will be same so subtract ways when both are together, that is (hH)T and T(Hh) => 6-2=4 ways
3) 3 heads hTT - so 3C1 = 3 ways
Total 1+4+3=8 ways

Total ways = 2^5=32

ans = 8/32=1/4
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Re: A fair coin is tossed 5 times.   [#permalink] 06 Jan 2019, 08:40
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A fair coin is tossed 5 times.

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