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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