You need a formula for this.
If the probability (in one toss) of getting head is "p", then the probability of getting exactly k heads out of n tosses is:
(n,k) p^k (1-p)^(n-k)
where (n,k) denotes the number of combinations of k elements among n.
You should also know that (n,k) = n! / (( n-k)! k! )
so here, with n=8, k=6, and p=.5 you have
(n,k) = 8*7 / 2 = 28
and your probability is :
28 * 1/2^6 * 1/2^2 = 28 / 256 = 7 / 64
There are eight possible outcomes: HHH, HHT, HTT, HTH, TTT, TTH, THH, THT. Of these, 3 contain two tails: HTT, TTH, and THT and the probability of getting two tails is 3/8. If the question were 'getting at least two tails' then TTT would need to be included for a probability of 4/8 or 0.5.
the answer is 2/13
There is a one in two chance of tails every time, rewritten as 1/2 The probability of tail 3 times in a row is 1/2 x 1/2 x 1/2= 1/8. So the answer is one in eight, or 12.5%
The probability of obtaining 7 heads in eight flips of a coin is:P(7H) = 8(1/2)8 = 0.03125 = 3.1%
The probability of getting at least 1 tails is (1 - probability of getting all heads) The probability of getting all heads (no tails) is ½ x ½ x ½ x ½ x ½ x ½ x ½ x ½ = 1/256 = 0.00390625 so the probability of getting at least ONE tails is 1-0.30390625 = 0.99609375 = 255/256
The probability of getting 8 heads out of 10 tosses is (10C8)(1/2)^8 (1/2)^2 = 45 / 1024 = 0.0439. It is assumed that the probability of getting a head in a single toss is 1/2. 10C8 = 10 x 9 / 1 x 2 = 45
The probability of getting exactly seven tails if you flip a coin eight times is: P(7T1H) = 8∙(1/2)8 =0.03125 ≈ 3.1%
There are eight possible outcomes: HHH, HHT, HTT, HTH, TTT, TTH, THH, THT. Of these, 3 contain two tails: HTT, TTH, and THT and the probability of getting two tails is 3/8. If the question were 'getting at least two tails' then TTT would need to be included for a probability of 4/8 or 0.5.
the answer is 2/13
(8!/(6!2!) / 2**8 = 28/256 = 7/64 = 0.109375
Probability that it is one of these eight cards is 8/52. Hence the probability of not getting these eight cards is 44/52
There is a one in two chance of tails every time, rewritten as 1/2 The probability of tail 3 times in a row is 1/2 x 1/2 x 1/2= 1/8. So the answer is one in eight, or 12.5%
Eight.
The probability of obtaining 7 heads in eight flips of a coin is:P(7H) = 8(1/2)8 = 0.03125 = 3.1%
The probability of getting at least 1 tails is (1 - probability of getting all heads) The probability of getting all heads (no tails) is ½ x ½ x ½ x ½ x ½ x ½ x ½ x ½ = 1/256 = 0.00390625 so the probability of getting at least ONE tails is 1-0.30390625 = 0.99609375 = 255/256
It is approx 0.1445
It is 93/256 = 0.363 approx.