The probability of 10 heads in a row is (0.5)10 = 0.000977 = 0.0977% .
It makes no difference what has come before or what comes after.
The answer depends on how many times the coin is tossed. The probability is zero if the coin is tossed only once! Making some assumptions and rewording your question as "If I toss a fair coin twice, what is the probability it comes up heads both times" then the probability of it being heads on any given toss is 0.5, and the probability of it being heads on both tosses is 0.5 x 0.5 = 0.25. If you toss it three times and want to know what the probability of it being heads exactly twice is, then the calculation is more complicated, but it comes out to 0.375.
Pr(H at least once in 10 tosses) = 1 - pr(No H in 10 tosses) = 1 - Pr(10 T in 10 tosses) = 1 - (1/2)10 = 1 - 1/1024 = 1023/1024
This is a good bet to take. Your expected payout is 0.5 each round of the 2 tosses. The possible outcomes from 2 tosses: HT, HH, TT, TH. The probability that heads comes up is 3 in 4 (.75). The probability that heads does not come up is 1 in 4 (.25). Your expected payout is: (2 * .75) + (-4 * .25) = 1.5 - 1 = 0.5
The chance is 3.5/7.
1/2
7/10
It is (1/2)9 = 1/512
0.5, 1/2, 50% The probability for heads versus tails does not change based on the amount of times the coin is tossed. It will always be a 50% chance.
it is a fair chance so 1/2 :P
Assuming a fair 6 sided die, there is a 1/6 probability for each number you could choose.
Heads comes up half the time, sis one time in six tosses so the combined odds of heads and six is 1/2 x 1/6 = 1/12, one time in 12.
The probability is that it comes out 7 times out of 10 tries, or 70% of the times.