Each coin toss is either a head or tail. The tosses done before have no impact on the toss going on.
There are always (mathematical speaking) a 50-50 chance of head or tail.
If you had 100 tails in a row doesn't matter. It's still 50-50.
The prob. of head in 6th (or 57th or whatever) toss is 50%.
The probability of a coin landing on heads is 0.5. It does not matter which toss it is, and it does not matter what the toss history was.
the probability of getting heads-heads-heads if you toss a coin three times is 1 out of 9.
No, when you toss a coin there is a 50 percent chance it will land heads up.
1/2, or 50% since you are only asking what the probability of the last outcome is.
Each independent trial has a 1/2 probability that a heads will result. So for a sequence of 4 heads you have 1/2 x 1/2 x 1/2 x 1/2 = 1/16
the probability is actually not quite even. It would actually land heads 495 out of 1000 times because the heads side is slightly heavier
Each coin toss is either a head or tail. The tosses done before have no impact on the toss going on. There are always (mathematical speaking) a 50-50 chance of head or tail. If you had 100 tails in a row doesn't matter. It's still 50-50. The prob. of head in 6th (or 57th or whatever) toss is 50%.
there is a 1/2 chance that it will land on heads and 1/2 chance it will land on tails. it dosent matter what the stats are
There is a 50% chance that it will land on heads each toss. You need to clarify the question: do you mean what is the probability that it will land on heads at least once, exactly once, all five times?
Assuming each toss is independent, you can use the binomial distribution,P( X = 32) = 50C32*(p)32*(1-p)50-32 where p is the probablity of getting heads on a single toss. Assuming that the coin is fair, p = 1/2.So the answer is 50C32*(1/2)50 = 50!/(32!*18!*250 = 0.016 approx.Assuming each toss is independent, you can use the binomial distribution,P( X = 32) = 50C32*(p)32*(1-p)50-32 where p is the probablity of getting heads on a single toss. Assuming that the coin is fair, p = 1/2.So the answer is 50C32*(1/2)50 = 50!/(32!*18!*250 = 0.016 approx.Assuming each toss is independent, you can use the binomial distribution,P( X = 32) = 50C32*(p)32*(1-p)50-32 where p is the probablity of getting heads on a single toss. Assuming that the coin is fair, p = 1/2.So the answer is 50C32*(1/2)50 = 50!/(32!*18!*250 = 0.016 approx.Assuming each toss is independent, you can use the binomial distribution,P( X = 32) = 50C32*(p)32*(1-p)50-32 where p is the probablity of getting heads on a single toss. Assuming that the coin is fair, p = 1/2.So the answer is 50C32*(1/2)50 = 50!/(32!*18!*250 = 0.016 approx.
Most likely 5,000 times.
No, not if it is a fair coin.