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
