Thank you this was helpful

Good solution

I have a little trick that helps prevent getting lost in repetition when listing HHHTT, THHHT.. and what not..
I just replace 3 heads in a row with A.
then my list of outcomes looks like this
ATT
TAT
TTA
HTA
ATH
THA
AHT
AHH

I hope this helps save some time and headache

Awesome trick, really helpful

Ofcourse, we could use a mix of combinations and logic too..

1) all 5 heads - 1 HHHHH
2) 4 heads - ways 5C4=5... Only one where T is in centre should be discarded HHTHH so 5-1=4
2) 3 heads - let 3 heads be one H so HTT, THT and TTH - so 3 ways
total - 1+4+3=8

Also the trick of Happymathtutor is good.
I would use this in combinations too solve..
Take all H to be 1 as h
1) all 5 heads - hHH - 1 way
2) 4 heads hHT - so ways = 3*2*1=6 but hH and Hh will be same so subtract ways when both are together, that is (hH)T and T(Hh) => 6-2=4 ways
3) 3 heads hTT - so 3C1 = 3 ways
Total 1+4+3=8 ways

Total ways = 2^5=32

ans = 8/32=1/4

Given that A fair coin is tossed 5 times and we need to find What is the probability of getting at least three heads on consecutive tosses

At least 3 heads means that we can get 3 or more heads in consecutive tosses.

=> P(At least 3 consecutive heads) = P(3 consecutive heads) + P(4 consecutive heads) + P(5 consecutive heads)

P(3 consecutive heads)

Total number of cases = \(2^5\) = 32
Case in which we get 3 consecutive heads are HHHTT, HHHTH, THHHT, TTHHH, HTHHH => 5

=> P(3 consecutive heads) = \(\frac{5}{32}\)

P(4 consecutive heads)

Case in which we get 4 consecutive heads are HHHHT, THHHH => 2

=> P(4 consecutive heads) = \(\frac{2}{32}\)

P(5 consecutive heads)

Case in which we get 5 consecutive heads are HHHHH => 1

=> P(5 consecutive heads) = \(\frac{1}{32}\)

=> P(At least 3 consecutive heads) = P(3 consecutive heads) + P(4 consecutive heads) + P(5 consecutive heads) = \(\frac{5}{32}\) + \(\frac{2}{32}\) + \(\frac{1}{32}\) = \(\frac{8}{32}\) = \(\frac{1}{4}\)

So, Answer will be B
Hope it helps!

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

First just consider how many ways you can get at least three consecutive Hs:

HHHTT
HHHHT (including the weird little ones HHHTH and HTHHH--remember you only need three in a row)
HHHHH

Then consider how many ways each of these can be organized to retain the three consecutives!

HHHTT or THHHT or TTHHH

HHHHT or THHHH or HHHTH or HTHHH

HHHHH only

That's eight possibilities total.

Now... how many independent possibilities do we have? There are two possibilities per turn, so that's 2^5 = 32 total possibilities.

So desired outcomes are 8 over total outcomes of 32: 8/32 = 1/4.

Answer B.

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