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Answered 2012-08-26 11:03:14

7*(1/2)7 = 7/128 = 5.47% approx.

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What is the probability of obtaining exactly seven heads in eight flips of a coin given that at least one is a head?

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


What is the probability of obtaining exactly six heads in seven flips of a coin given that at least one is a head?

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.


What is the probability of obtaining exactly four heads in five flips of a coin given that at least three are heads?

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.


What is the probability of obtaining exactly two heads in three flips of a coin given that at least one is a head?

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


What is the probability of exactly three heads in four flips of a coin given at least two are heads?

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%.



What is the probability of obtaining exactly 4 heads in a coin?

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


What is the probability of obtaining exactly three heads in four flips of a coin given that at least two are heads?

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



What are the odds against getting exactly two heads in three successive flips of a coin?

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.


What is the probability of obtaining heads and a two?

The answer depends on what the experiment is!


What is the probability of obtaining exactly three heads in four flips of a coin given that at least one is a head?

50-50. each toss is independent of any previous toss. if all tosses are to be heads/tails then each toss you multiply by the number of chances. i,e. 2, starting with 1. three heads in a row is 1x2x2




What is the probability of obtaining exactly five heads in six flips of a coin given that at least one is a head?

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.


What is the probabilty that a coin will land on heads if flipped 8 times?

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.



What is the probabity of obtaining three heads?

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))


Why is it called experimental probability?

It's difficult to think of a real event to which an exact probability can be assigned. We say that flipping a coin yields 'heads' with probability 1/2 but we do not know that definitely. The only way of assigning a probability in the sense of numbers of heads versus total numbers of flips is by experiment. (Be aware though that there are other interpretations of the word probability.) If I were to flip a coin 500 times and obtained 249 heads then the experimental probability of obtaining a head would be 249/500 or 0.498.


Probability to get exactly 2 heads when flipping a coin 3 times?

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.



What is the probability of flipping a coin twice and getting two heads?

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.


What is the expected number of flips of a coin to simulate a six-sided die?

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.



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