Yes you are right. I must take care next time

Corrected

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

Other than putting the series on paper, is there is a better way to answer this question?

I’m also wondering like @mohan514 if there’s another place way?

No. I think no other way is possible or a shortcut.

Maybe @GreenlightTestPrep has more approaches.

Regards

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

nope, because you need to get exaclty 3 heads, no more than that.

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

Consider how many ways you can get EXACTLY three in a row. That is, no more or no less than three together. We can have four or five Hs, but we have to make sure that only the three are in a row (six Hs will be out because that's clearly not only three in a row).

3 and 3: HHHTTT or TTTHHH or THHHTT or TTHHHT

4 and 2: HHHTHT or HHHTTH or THTHHH or HTTHHH or THHHTH or HTHHHT

5 and 1: HHHTHH or HHTHHH

That's 12 total possibilities where the Hs lie exactly three in a row.

Given that we have six flips, that means we have 2^6 = 64 total possibilities.

Desired outcomes is 12 over total outcomes of 64, or 12/64 = 3/16.

As for the argument about HHHHHH, remember, that HHHHHH is categorically NOT "exactly three heads in a row." The word exactly in GRE terms functions as an "if and only if" in logic: three and only three consecutive heads count. Four consecutives is excluded. Five consecutives is excluded. Six consecutives is excluded.

It is important to be totally precise with the wording on questions like these as this can, as we're seeing, lead to troublesome misinterpretations.

Answer

Hey, if the problem states "exactly 3 heads in a row, why are we considering 4H's? shouldn't we just consider these?
HHHTTT
THHHTT
TTHHHT
TTTHHH

No, the wording suggests that we need to consider any case that has three and only three heads in a row. It is irrelevant whether another head is outside the group of three.

HHHTHT or HHHTHH, for example, have three and only three in a row despite the straggling Hs that are not in that group of three. These cases still count.

The question wording does not suggest "only three heads total out of six."

NiharikaG if the problem states "exactly 3 heads in a row"

Exactly three heads in a row does not exclude the option where we are having Heads after a space of one T.

Example: H H H T XX is valid combination where X can be anything out of T or H
HHHTXX just makes sure that the combination of 3 heads HHH is either in the starting or the end and is preceded and followed by at least one Tail

Similarly, THHHTX, XTHHHT, XXTHHH are also valid combinations.

Hope it helps!