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If a fair coin is tossed six times, what is the probability

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If a fair coin is tossed six times, what is the probability [#permalink]  04 Oct 2017, 20:33
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If a fair coin is tossed six times, what is the probability of getting exactly three heads in a row?

(A) 1/16
(B) 3/16
(C) 1/8
(D) 3/8
(E) 1/2

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[Reveal] Spoiler: OA
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Re: If a fair coin is tossed six times, what is the probability [#permalink]  04 Oct 2017, 22:18
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Bunuel wrote:
If a fair coin is tossed six times, what is the probability of getting exactly three heads in a row?

(A) 1/16
(B) 3/16
(C) 1/8
(D) 3/8
(E) 1/2

Kudos for correct solution.

The total number of outcomes = 2 ^6 =64 (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 exactly 3 heads .

So Favorable outcomes E ={3 heads and remaining 3 tails, because it says exactly 3 heads}
=H H H T H T
H H H T T H
H H H T H H
H T H H H T
H T T H H H
H H T H H H
T H H H T T
T H H H T H
T H T H H H
T T H H H T
T T T H H H

So no. of favorable outcomes E =12
So the required probability = $$\frac{12}{64}$$ = $$\frac{3}{16}$$
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Re: If a fair coin is tossed six times, what is the probability [#permalink]  04 Oct 2017, 22:20
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pranab01 wrote:
Bunuel wrote:
If a fair coin is tossed six times, what is the probability of getting exactly three heads in a row?

(A) 1/16
(B) 3/16
(C) 1/8
(D) 3/8
(E) 1/2

Kudos for correct solution.

The total number of outcomes = 2 ^6 =64 (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 exactly 3 heads .

So Favorable outcomes E ={3 heads and remaining 3 tails, because it says exactly 3 heads}
=H H H T H T
H H H T T H
H H H T H H
H T H H H T
H T T H H H
H H T H H H
T H H H T T
T H H H T H
T H T H H H
T T H H H T
T T T H H H

So no. of favorable outcomes E =12
So the required probability = $$\frac{12}{64}$$ = $$\frac{3}{12}$$

Small typo there: 12/64 = 3/16, not 3/12.
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Re: If a fair coin is tossed six times, what is the probability [#permalink]  04 Oct 2017, 22:32
Bunuel wrote:

Small typo there: 12/64 = 3/16, not 3/12.

Yes you are right. I must take care next time

Corrected
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Re: If a fair coin is tossed six times, what is the probability [#permalink]  07 Apr 2018, 09:15
pranab01 wrote:
Bunuel wrote:
If a fair coin is tossed six times, what is the probability of getting exactly three heads in a row?

(A) 1/16
(B) 3/16
(C) 1/8
(D) 3/8
(E) 1/2

Kudos for correct solution.

The total number of outcomes = 2 ^6 =64 (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 exactly 3 heads .

So Favorable outcomes E ={3 heads and remaining 3 tails, because it says exactly 3 heads}
=H H H T H T
H H H T T H
H H H T H H
H T H H H T
H T T H H H
H H T H H H
T H H H T T
T H H H T H
T H T H H H
T T H H H T
T T T H H H

So no. of favorable outcomes E =12
So the required probability = $$\frac{12}{64}$$ = $$\frac{3}{16}$$

If you included H H H T H T, why are you not including H H H H H H ? there're exactly 3 Head consecutively in that arrangement. The answer is E
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Re: If a fair coin is tossed six times, what is the probability [#permalink]  07 Apr 2018, 23:27
Other than putting the series on paper, is there is a better way to answer this question?
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Re: If a fair coin is tossed six times, what is the probability [#permalink]  08 Apr 2018, 05:09
I’m also wondering like @mohan514 if there’s another place way?
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Re: If a fair coin is tossed six times, what is the probability [#permalink]  08 Apr 2018, 10:34
Expert's post
No. I think no other way is possible or a shortcut.

Maybe @GreenlightTestPrep has more approaches.

Regards
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Re: If a fair coin is tossed six times, what is the probability [#permalink]  23 Sep 2018, 23:17
See there are 6 boxes to fill.
For such probs. always imagine things..try to visualize
If you fill first three by H (which stands for Head), only the possibility of 2 Hs exist that too after a gap of 1 box: HHHTHH== Max. 5 Hs expected
Now, similarly second way in which 3 Hs can come in a row: THHHTH==Thus only 4 Hs expected at max
similarly Third way: HTHHHT=4 Hs again at max
now Fourth way: HHTHHH= 5 Hs at max.

So do sum of individual possibilities of H in all 4 instances:
1: (1/2)^3 * (1/2)^2
2: (1/2)^3 * (1/2)
3: (1/2)^3 * (1/2)
4: (1/2)^3*(1/2)^2
Calculating===> (1/2)^4 {1/2+1+1+1/2}
===> 1/16*6/2===>3/16

Hope that Helps.

PS in probablities, there r no shortcuts please dont go by formulas rather try visualize imho
Re: If a fair coin is tossed six times, what is the probability   [#permalink] 23 Sep 2018, 23:17
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