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Answered 2011-08-08 06:05:37

About a 1 in 16 chance of getting a coin to land on heads 4 times in a row.

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The probability of obtaining 7 heads in eight flips of a coin is:P(7H) = 8(1/2)8 = 0.03125 = 3.1%

The requirement that one coin is a head is superfluous and does not matter. The simplified question is "what is the probability of obtaining exactly six heads in seven flips of a coin?"... There are 128 permutations (27) of seven coins, or seven flips of one coin. Of these, there are seven permutations where there are exactly six heads, i.e. where there is only one tail. The probability, then, of tossing six heads in seven coin tosses is 7 in 128, or 0.0546875.

We can simplify the question by putting it this way: what is the probability that exactly one out of two coin flips is a head? Our options are HH, HT, TH, TT. Two of these four have exactly one head. So 2/4=.5 is the answer.

The probability of obtaining exactly two heads in three flips of a coin is 0.5x0.5x0.5 (for the probabilities) x3 (for the number of ways it could happen). This is 0.375. However, we are told that at least one is a head, so the probability that we got 3 tails was impossible. This probability is 0.53 or 0.125. To deduct this we need to divide the probability we have by 1-0.125 0.375/(1-0.125) = approximately 0.4286

If it is a fair coin, the probability is 1/4.If it is a fair coin, the probability is 1/4.If it is a fair coin, the probability is 1/4.If it is a fair coin, the probability is 1/4.

The probability of the coin flip being heads or tails is 100%.

The probability that you will toss five heads in six coin tosses given that at least one is a head is the same as the probability of tossing four heads in five coin tosses1. There are 32 permutations of five coins. Five of them have four heads2. This is a probability of 5 in 32, or 0.15625. ----------------------------------------------------------------------------------- 1Simplify the problem. It asked about five heads but said that at least one was a head. That is redundant, and can be ignored. 2This problem was solved by simple inspection. If there are four heads in five coins, this means that there is one tail in five coins. That fact simplifies the calculation to five permutations exactly.

The probability of flipping Heads on a coin is 1 - a certainty - if the coin is flipped often enough. On a single toss of a fair coin the probability is 1/2.

The probability of getting 3 or more heads in a row, one or more times is 520/1024 = 0.508 Of these, the probability of getting exactly 3 heads in a row, exactly once is 244/1024 = 0.238

(7!/(4!*3!))*(1/2)^3 * (1/2)^4 = 35*(1/2)^7 which is about 0.2734375

If it is a fair coin, the probability is exactly 50%. The coin has no memory of what it did in the last flip. ■

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.

The probability of throwing exactly 2 heads in three flips of a coin is 3 in 8, or 0.375. There are 8 outcomes of flipping a coin 3 times, HHH, HHT, HTH, HTT, THH, THT, TTH, and TTT. Of those outcomes, 3 contain two heads, so the answer is 3 in 8.

The probability of tossing a coin and getting heads is 0.5

If it is a fair coin then the probability is 0.5

If it is a fair coin, the probability is 1/2.

The probability to tossing a coin and obtaining tails is 0.5. Rolling a die has nothing to do with this outcome - it is unrelated.

The probability that both coins are heads is the probability of one coin landing heads multiplied by the probability of the second coin landing heads: (.5) * (.5) = .25 or (1/2) * (1/2) = 1/4

The empirical probability can only be determined by carrying out the experiment a very large number of times. Otherwise it would be the theoretical probability.

Pr(3H given >= 2H) = Pr(3H and >= 2H)/Pr(>=2H) = Pr(3H)/Pr(>=2H) = (1/4)/(11/16) = 4/11.

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%.

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