3 to 1
Three in eight are the odds of getting exactly two heads in three coin flips. There are eight ways the three flips can end up, and you can get two heads and a tail, a head and a tail and a head, or a tail and two heads to get exactly two heads.
1/10 not likely
It's human nature to think so, but no. Even after getting 100 heads in a row, the odds of the next flip being heads is still 50%.
Five coin flips. Any outcome on a six-sided die has a probability of 1 in 6. If I assume that the order of the outcome does not matter, the same probability can be achieved with five flips of the coin. The possible outcomes of five flips of a coin are as follows: 5 Heads 5 Tails 4 Heads and 1 Tails 4 Tails and 1 Heads 3 Heads and 2 Tails 3 Tails and 2 Heads For six possible outcomes.
If you know that two of the four are already heads, then all you need to find isthe probability of exactly one heads in the last two flips.Number of possible outcomes of one flip of one coin = 2Number of possible outcomes in two flips = 4Number of the four outcomes that include a single heads = 2.Probability of a single heads in the last two flips = 2/4 = 50%.
Three in eight are the odds of getting exactly two heads in three coin flips. There are eight ways the three flips can end up, and you can get two heads and a tail, a head and a tail and a head, or a tail and two heads to get exactly two heads.
1/10 not likely
0.53 = 0.125.
For 3 coin flips: 87% chance of getting heads at least once 25% chance of getting heads twice 13% chance of getting heads all three times
On a fair 50-50 coin, the chance of you getting heads 3 times in a row is .5 * .5 * .5 which is 12.5% Getting exactly 3 heads out of any number of coin flips involves: (Number of Flips!/ [6 * (N-3)!]) * (.5^3)* (.5^(n-3))
1/16
It's human nature to think so, but no. Even after getting 100 heads in a row, the odds of the next flip being heads is still 50%.
The probability is 25%. The probability of flipping a coin once and getting heads is 50%. In your example, you get heads twice -- over the course of 2 flips. So there are two 50% probabilities that you need to combine to get the probability for getting two heads in two flips. So turn 50% into a decimal --> 0.5 Multiply the two 50% probabilities together --> 0.5 x 0.5 = 0.25. Therefore, 0.25 or 25% is the probability of flipping a coin twice and getting heads both times.
Five coin flips. Any outcome on a six-sided die has a probability of 1 in 6. If I assume that the order of the outcome does not matter, the same probability can be achieved with five flips of the coin. The possible outcomes of five flips of a coin are as follows: 5 Heads 5 Tails 4 Heads and 1 Tails 4 Tails and 1 Heads 3 Heads and 2 Tails 3 Tails and 2 Heads For six possible outcomes.
50%
This is a probability question. Probabilities are calculated with this simple equation: Chances of Success / [Chances of Success + Chances of Failure (or Total Chances)] If I flip a coin, there is one chance that it will land on heads and one chance it will land on tails. If success = landing on heads, then: Chances of Success = 1 Chances of Failure = 1 Total Chances = 2 Thus the probability that a coin will land on heads on one flip is 1/2 = .5 = 50 percent. (Note that probability can never be higher than 100 percent. If you get greater than 100 you did the problem incorrectly) Your question is unclear whether you mean the probability that a coin will land on head on any of 8 flips or all of 8 flips. To calculate either you could write out all the possible outcomes of the flips (for example: heads-heads-tails-tails-heads-tails-heads-heads) but that would take forvever. Luckily, because the outcome of one coin flip does not affect the next flip you can calculate the total probability my multiplying the probabilities of each individual outcome. For example: Probability That All 8 Flips Are Heads = Prob. Flip 1 is Heads * Prob. Flip 2 is Heads * Prob. Flip 3 is Heads...and so on Since we know that the probability of getting heads on any one flips is .5: Probability That All 8 Flips Are Heads = .5 * .5 * .5 * .5 * .5 * .5 * .5 * .5 (or .58) Probability That All 8 Flips Are Heads = .00391 or .391 percent. The probability that you will flip a heads on any of flips is similar, but instead of thinking about what is the possiblity of success, it is easier to approach it in another way. The is only one case where you will not a heads on any coin toss. That is if every outcome was tails. The probability of that occurring is the same as the probability of getting a heads on every toss because the probability of getting a heads or tails on any one toss is 50 percent. (If this does not make sense redo the problem above with tails instead of heads and see if your answer changes.) However this is the probability of FAILURE not success. This is where another probability formula comes into play: Probability of Success + Probability of Failure = 1 We know the probability of failure in this case is .00391 so: Probability of Success + .00391 = 1 Probability of Success = .9961 or 99.61 percent. Therefore, the probability of flipping a heads at least once during 8 coin flips is 99.61 percent. The probability of flipping a heads every time during 8 coin flips is .391 percent.
If you know that two of the four are already heads, then all you need to find isthe probability of exactly one heads in the last two flips.Number of possible outcomes of one flip of one coin = 2Number of possible outcomes in two flips = 4Number of the four outcomes that include a single heads = 2.Probability of a single heads in the last two flips = 2/4 = 50%.