revised, but not sure

if we consider 10 tosses with two clued (consecutive) attempts resulting in heads, then we have (10-1) consecutive attempts with heads out of 2^10 total possibilities <> 2*2*2 ... each time two options can be effective throughout ten attempts

two consecutive and eight other attempts are randomly chosen out of two options (head-tail), hence (2^8) for each of 9 attempts

to find P(0 consecutive attempts heads)= 1 - (2^8)/2^10 = 1 -1/4 = 3/4

d

knight247 wrote: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
E)Can't be determined

This is extremely unrealistic and is not worth worrying about.

knight247 wrote: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
E)Can't be determined

Firstly, number of times 'heads' come should not be more than 5.( Otherwise, we will definetely have consecutive heads)

Number of ways when NO 'heads' come : 11c0 = 1
Number of ways when 1 'heads' come : 10c1 = 10
Number of ways when 2 'heads' come : 9c2 = 36
Number of ways when 3 'heads' come : 8c3 = 56
Number of ways when 4 'heads' come : 7c4 = 35
Number of ways when 5 'heads' come : 6c5 = 6

Total favourable outcome = 144
All possible = 2^10

Answer : 144/2^10 = 9/2^6 ( Option D. None of these )

Please explain this one.

Thanks
Mankey