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Q: Maria flips a dime four times What is the probability that she will get at least two heads?

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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%

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%.

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

If it is a fair coin, the probability of getting at least one Head from 3 flips is 7/8If it is a fair coin, the probability of getting at least one Head from 3 flips is 7/8If it is a fair coin, the probability of getting at least one Head from 3 flips is 7/8If it is a fair coin, the probability of getting at least one Head from 3 flips is 7/8

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.

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

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.

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.

As the question is "what is the probability of getting at least one head" the correct way to answer this is to ask what is the probability of not getting any heads and then subtract this from 1.The probability of not getting a head in 4 flips = 0.54 (i.e. 0.5 * 0.5 * 0.5 * 0.5) = 1/16.Therefore the probability of getting at least one head is 1 - 1/16 = 15/16.

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

Probability of no heads = (0.5)^5 = 0.03125Probability of at least one head = 1 - probability of no heads = 1 - 0.03125 = 0.96875

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