chauhan2011 wrote:

Bunuel ,

Could yo help with this one please.

A fair coin is tossed 10 times. What is the probability that two heads do not occur consecutively?

A) 1/ 2^4

B) 1/2^3

C)1/2^5

D) None of the above

This is very hard problem, way beyond the GMAT scope. It was discussed before and many solutions have been given. Here is mine:

Possible number of patterns (total number of combinations) 2^n (each time either H or T=2 outcomes, 10 times=2^10).

Let's check two consecutive H:

If we toss once we'll have 2^1=2 combinations:

H, T -

2 outcomes with NO 2 consecutive H.

If we toss twice we'll have 2^2=4 combinations:

HT, TH, TT, HH -

3 outcomes with NO 2 consecutive H.

If we toss 3 times we'll have 2^3=8 combinations:

TTT, TTH, THT, HTT, HTH, HHT, THH, HHH 5 outcomes with NO 2 consecutive H.

If we toss 4 times we'll have 2^4=16 combinations:...

8 outcomes with NO 2 consecutive H.

...

On this stage we can see the pattern in "no consecutive H": 2, 3, 5, 8...

I guess it's Fibonacci type of sequence and it will continue: 2, 3, 5, 8, 13, 21, 34, 55, 89,

144.

144 is outcomes with no consecutive H if we toss 10 times.

P(no two consecutive H in 10 toss)=144/2^10=144/1024=14.0625%

Other solutions and discussion at:

probability-coin-toss-57447.htmlHope it helps. BTW were did you find this question?

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