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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

my approach (1/2)^6 * 6C3 = 5/16 but the ans says it is b. where i am wrong??

6C3 gives the number of any 3 being heads. We need three heads in a row. Also, notice that in this scenario we can have more than 3 heads in total:
H H HTHT
H H HTTH
H H HTHH
H H HTTT

TH H HTT
TH H HTH

HTH H HT
TTH H HT

HTTH H H
HHTH H H
TTTH H H
THTH H H

but the question asks exactly three heads, doesnt that mean not more than 3 heads in a row?? is there any shortcut method since its totally irrational to write this scenaio in exam time.
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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

my approach (1/2)^6 * 6C3 = 5/16 but the ans says it is b. where i am wrong??

6C3 gives the number of any 3 being heads. We need three heads in a row. Also, notice that in this scenario we can have more than 3 heads in total:
H H HTHT
H H HTTH
H H HTHH
H H HTTT

TH H HTT
TH H HTH

HTH H HT
TTH H HT

HTTH H H
HHTH H H
TTTH H H
THTH H H

but the question asks exactly three heads, doesnt that mean not more than 3 heads in a row?? is there any shortcut method since its totally irrational to write this scenaio in exam time.

The question asks about exactly three heads in a row. No case above has more than 3 heads in a row.
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Would like to add my little bit:

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 :)
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Given that A fair coin is tossed six times, and we need to find what is the probability of getting exactly three heads in a row

Coin is tossed 6 times => Total number of cases = \(2^6\) = 64

We need to put three heads together starting from the first position and make sure that before and after three consecutive heads we have tails.

Following are the cases possible:

HHHTXX
THHHTX
XTHHHT
XXTHHH

where X be Tail or Head

=> Total Cases are
HHHTXX -> Each X can be T or H => 2*2 = 4 cases
THHHTX -> Single X can be T or H => 2 cases
XTHHHT -> Single X can be T or H => 2 cases
XXTHHH -> Each X can be T or H => 2*2 = 4 cases

Total cases = 4 + 2 + 2 + 4 = 12

=> Probability(Exactly 3 heads in a row) = \(\frac{12}{64}\) = \(\frac{3}{16}\)

So, Answer will be B
Hope it helps!

Watch the following video to learn How to Solve Probability with Coin Toss Problems

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