VeritasPrepKarishma wrote:

ankitranjan wrote:

If an unbiased coin is flipped 5 times,the probabilty that the same face does not show up in any three consecutive flips is

A. 1/2

B. 5/8

C. 3/8

D. 7/8

E. 9/8

I agree with Bunuel. The easiest solution I could think of was enumerating in a sequence. I try and find how I can have the same face three or more times.

Start with 3 Hs and 2 Ts, go on to 4Hs and 1T and finally 5Hs

The 3 Hs have to be together so: HHHTT, TTHHH, THHHT

Then, 4 Hs and 1T. Place the 4 Hs first: H H H H. Now look for the places where you can place the T: THHHH, HTHHH, HHHTH and HHHHT

Finally 5 Hs: HHHHH

Total 8 ways to get 3 or more Heads in a row. Ways to get 3 or more Tails in a row is also 8.

Total number of ways outcomes with 5 coins = 2^5 = 32

Probability of same face occurring 3 or more times = 16/32 = 1/2

Probability of same face occurring less than 3 times = 1 - 1/2 = 1/2

Is there a straighter way to solve this problem? Without going the opposite way

My attempt: Of the five slots, three straight slots are reserved for opposite results. (HTH, THT). The remaining two slots can yield a head or a tail.

So the possible number of outcomes is 2*2 = 4. Now these two slots are always clubbed together and can be treated as one slot that can be adjusted anywhere among the 4 possible slots in 4c1 ways. So the total possible outcomes under the given conditions will be 2*2* 4c1 = 16

Probability is 16/32 = 1/2

Is the thinking behind reaching this solution correct? Or did I just get lucky with the answer?

TIA

KC