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Director  Status: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing.
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A fair coin is tossed 5 times. What is the probability of getting at  [#permalink]

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Difficulty:   95% (hard)

Question Stats: 44% (02:20) correct 56% (02:27) wrong based on 611 sessions

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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

Originally posted by nusmavrik on 20 Aug 2010, 05:39.
Last edited by Bunuel on 05 Feb 2019, 05:21, edited 1 time in total.
Renamed the topic.
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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
12
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

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

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

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Director  Status: Impossible is not a fact. It's an opinion. It's a dare. Impossible is nothing.
Affiliations: University of Chicago Booth School of Business
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Re: A fair coin is tossed 5 times. What is the probability of getting at  [#permalink]

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Awesome explanation Bunuel ! Manager  Joined: 27 May 2010
Posts: 158
Re: A fair coin is tossed 5 times. What is the probability of getting at  [#permalink]

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2
Brunnel explained it in detail. I would just count the number of possibilities and divide it by 2^n

HHHHH
HHHTT
HHHHT
THHHT
TTHHH
HHHTH
THHHH
HTHHH

8/(2)^5 ==> 8/32 ==> 1/4
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Re: A fair coin is tossed 5 times. What is the probability of getting at  [#permalink]

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All possible events = (1/2)^5= 1/32
all the favorable events
HHHHH
HHHTT
THHHT
TTHHH
HHHHT
THHHH
that makes 6 favourable events
Hence answer will 6/32 = 1/4 therefore B
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Re: A fair coin is tossed 5 times. What is the probability of getting at  [#permalink]

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my question is how do you determine from this problem, that immediately, you are going to have to map out the possibilities?

there seems to be no nifty way to do it other than really just exhausting the possibilities?

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

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1
Bunuel - U rock.
+1 for you.

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

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This is like another probability question asked in the forum, bunnel explained that very well.
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Re: A fair coin is tossed 5 times. What is the probability of getting at  [#permalink]

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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. 3/16
B. 1/4
C. 7/24
D. 5/16
E. 15/32

Kaplan says the answer is 1/4. I think the answer is 3/16 because I have interpreted the qn as: What is the probability of getting atleast 3 consecutive heads -> 3 consecutive heads or 4 consecutive heads or 5 consecutive heads ?
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Re: A fair coin is tossed 5 times. What is the probability of getting at  [#permalink]

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shahlanuk wrote:
A fair coin is tossed 5 times. What is the probability of getting at least three heads on consecutive tosses?

A. 3/16
B. 1/4
C. 7/24
D. 5/16
E. 15/32

Kaplan says the answer is 1/4. I think the answer is 3/16 because I have interpreted the qn as: What is the probability of getting atleast 3 consecutive heads -> 3 consecutive heads or 4 consecutive heads or 5 consecutive heads ?

Merging similar topics.

Your interpretation is correct: at least 3 consecutive heads means 3, 4, or 5 consecutive heads, but the answer is still 1/4. Please check the solution provided above and ask if anything remains unclear.
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Re: A fair coin is tossed 5 times. What is the probability of getting at  [#permalink]

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bellcurve wrote:
A fair coin is tossed 5 times, what is the probability of getting at least 3 heads on consecutive tosses?

-- Ans will be provided later.

Favorable outcomes:

THHHH

THHHT
TTHHH
HTHHH
HHHTH

Total possiblities = 2^5 = 32
Favorable = 8
Probability = 8/32 = 1/4

Hope it helps!

PS: Please provide atleast options if not OA.
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Re: A fair coin is tossed 5 times. What is the probability of getting at  [#permalink]

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5 heads - HHHHH (1 way)
4 heads - THHHH, HHHHT (2 ways)
3 heads - HHHTH, HHHTT, THHHT, HTHHH, TTHHH (5 ways)

Total 8 ways

Total no ways irrespective of outcome= 2x2x2x2x2 = 32

probability of atleast 3 consecutive heads = 8/32 or .25
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Re: A fair coin is tossed 5 times. What is the probability of getting at  [#permalink]

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3
1
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 $$8/32 = 1/4.$$
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Re: A fair coin is tossed 5 times. What is the probability of getting at  [#permalink]

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I prefer using Binomial Expansion:

(H + T)^5: H^5 + 5H^4(T) + 10H^3*T^2 + 10H^2*T^3 + 5H*T^4 + T^5
The outcome for 3Heads 2 Tails = 10H^3*T^2 => 10*(1/32) = 5/16

The correct value should be 5/32

Can someone correct what am missing out here?

Thanks.
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GMAT 1: 800 Q51 V51 Re: A fair coin is tossed 5 times. What is the probability of getting at  [#permalink]

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1
gmatbull wrote:
I prefer using Binomial Expansion:

(H + T)^5: H^5 + 5H^4(T) + 10H^3*T^2 + 10H^2*T^3 + 5H*T^4 + T^5
The outcome for 3Heads 2 Tails = 10H^3*T^2 => 10*(1/32) = 5/16

The correct value should be 5/32

Can someone correct what am missing out here?

Yes. It's not just three heads two tails. Its three heads in a row, AND, it could also be four heads in a row or five heads in a row.
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Re: A fair coin is tossed 5 times. What is the probability of getting at  [#permalink]

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I missed two cases
HHHTH and HTHHH .

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

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Can someone suggest another way of solving this without writing down the possibilities? Maybe use combinations.An alternative could be useful in case you miss out on a case while jotting down the possibilities..thanks
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Re: A fair coin is tossed 5 times. What is the probability of getting at  [#permalink]

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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

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

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

Hello Bunuel,

Can we solve this question by combination formula? Hope I am not asking wrong question. (Like for 3 consecutive heads= 5!/(3!*2!) x (1/2)^5. Using this, I am not getting 5/32. Please explain. Thanks
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Re: A fair coin is tossed 5 times. What is the probability of getting at  [#permalink]

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In this question, i find "Counting" the cases is the best way to solve.

However, to answer the last question posted : In the combination (as is used by you for 3 consecutive heads) case, we are taking 3 heads in five tosses and not 3 consecutive heads in as many tosses.
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Re: A fair coin is tossed 5 times. What is the probability of getting at  [#permalink]

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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 ? Re: A fair coin is tossed 5 times. What is the probability of getting at   [#permalink] 15 May 2016, 20:31

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