Math and Arithmetic

Statistics

Probability

Top Answer

If you know that two of the four are already heads, then all you need to find is

the probability of exactly one heads in the last two flips.

Number of possible outcomes of one flip of one coin = 2

Number of possible outcomes in two flips = 4

Number 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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0The 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

you toss 3 coins what is the probability that you get exactly 2 heads given that you get at least one head?

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

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

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.

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.

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.

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

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.

We have no way of knowing the probability of any given person flipping any given coin at any given time. But for any two flips of an honest coin, the probability that both are tails is 25% . (1/4, or 3 to 1 against)

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.

The probability is 0.375

One out of every four flips

The probability that exactly one will land heads up is 0.15625

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

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 a fair coin to land head is 1/2. Since for 4 flips it must land heads each time, the probability of 4 heads is 1/2 * 1/2 * 1/2 * 1/2 = 1/16.

(1/2)*(1/2)*(1/2)*(1/2)*(1/2)*(1/2) = 1/26 = 1/64

Pr(3 flips at least one H) = 1 - Pr(3 flips, NO heads) = 1 - Pr(3 flips, TTT) = 1 - (1/2)3 = 1 - 1/8 = 7/8

There is a 1/8 chance to land three heads.

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