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

Hi,
Just a bit of observation from my side..

1) why (1/2)^6 * 6C3 = 5/16 is wrong?

the formula you have used is for the ways 6 tosses will have exactly 3 heads...
so..A) you will have a combination like THTHTH, which cannot be included as 3 heads are not in a row..
B) you will not have a combination HHHTTH, as this has 4 heads overall but fits in in the restriction that there are 3 heads in a row..

2) then what is the way

Bunuel has ofcourse given it, but again ill share some info on it..
For 3 heads in a row, we have to have atleast 3 heads, so lets work out ways..
A) 3 heads..
all three heads can be taken as one, so total are 4- 3H,T,T,T...
3H can take any of four places= 4 ways..

B) 4 Heads..
again 3H are one but it cannot have fourth one next to it..
(i)so if we have 3H in edges.. 3H,T,_,_
the last two can be H,T or T,H--- so 2 ways..
similarily if 3H in the rightmost position-- _,_,T,3H---- 2 ways..

(ii) if we have 3H in middle... T,3H,T,H and H,T,3H,T--- 2ways..
TOTAL=2+2+2=6 ways..

C) 5 heads..
Again 3H in edges-- 3H,T,H,H or H,H,T,3H--2 ways.
we cannot have 3H in middle, as it will surely have a H adjacent to it, resulting in 4 H in a row..

D) 6 Heads--
NO possiblity..

TOTAL ways to have exactly 3H in a row=4+6+2=12 ways..
total combinations of 6 tosses= 2^6..
Probability= 12/2^6= 3/16


B
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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
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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

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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