Difficult Math Problem #85 - Probability

Using the bernoulli distribution, lets add p(getting 7 tails) + p(getting 8 tails):

(8C7)*(0.5^7)*(0.5) + (8C8)*(0.5^8) = 8/(2^8) + 1/(2^8) = 9/256

It's also the same as the probability of getting at most 1 head. We calculate p(at most 1 head) = p(0 heads) + p(1 head):

(8C0)*(0.5^0)*(0.5^8) + (8C1)*(0.5^1)*(0.5^7) = 1/(2^8) + 8/(2^8) = 9/256

OA

Number of ways 7 tails can turn up = 8C7
the probability of those is 1/2 each
Since the question asks for at least 7, we need to find the prob of all 8 tails - the number of ways is 8C8 = 1
Add the two probabilities
8C7*(1/2)^8 = 8/2^8 -- for getting 7
Prob of getting 8 tails = 1/2^8
Total prob = 8/2^8+1/2^8 = 9/2^8

Ans is E.

i think i am being asked for probability of 7 tails & probability of getting 8 tails in 8 throws
total possible outcomes=2(possible outcomes on any throw)*8(total throws)
=16

probability that he gets tails on at least 7 of the 8 flips=7/16*8/16=7/32
but this is wrong answer. can someone suggest how to calculate the right answer without combinatorics or suggest how to use Bernoulli distribution (i do not understand whether it is used to combine probabilities or it is used after we calculate total possible and favorable events)

Ashish, others,

Answer without formulas:

Totals 8 flips: possible arrangements = 2^8

we need 7 tails and 8 tails
in 8 ways we can have 7 tails [as 1 h can be at 8 places]
and
in 1 way we can have 8 tails

so favorable outcomes = 8+1 = 9
Total outcomes = 2^8 = 256

Where can I find
Source: Difficult Math Problems doc on this forum

Thanks