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

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It is approx 0.2461

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.

Assume the coin is fair, so there are equal amount of probabilities for the choices.There are two possible choices for a flip of a fair coin - either a head or a tail. The probability of getting a head is ½. Similarly, the probability of getting a tail is ½.Use Binomial to work out this problem. You should get:(5 choose 4)(½)4(½).(5 choose 4) indicates the total number of ways to obtain 4 tails in 5 flips.(½)4 indicates the probability of obtaining 4 tails.(½) indicates the probability of obtaining the remaining number of head.Therefore, the probability is 5/32.

The probability is 1/2 if the coin is flipped only twice. As the number of flips increases, the probability approaches 1.

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It is approx 0.2461

It is approx 0.0938

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

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

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.

Assume the coin is fair, so there are equal amount of probabilities for the choices.There are two possible choices for a flip of a fair coin - either a head or a tail. The probability of getting a head is ½. Similarly, the probability of getting a tail is ½.Use Binomial to work out this problem. You should get:(5 choose 4)(½)4(½).(5 choose 4) indicates the total number of ways to obtain 4 tails in 5 flips.(½)4 indicates the probability of obtaining 4 tails.(½) indicates the probability of obtaining the remaining number of head.Therefore, the probability is 5/32.

1/2 * 1/2 * 1/2 = 1/8 = 12.5%

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 is 1/2 if the coin is flipped only twice. As the number of flips increases, the probability approaches 1.

Assume the coin is fair, so there are equal amount of probabilities for the choices.There are two possible choices for a flip of a fair coin - either a head or a tail. The probability of getting a head is Â½. Similarly, the probability of getting a tail is Â½.Use Binomial to work out this problem. You should get:(5 choose 4)(Â½)4(Â½).(5 choose 4) indicates the total number of ways to obtain 4 tails in 5 flips.(Â½)4 indicates the probability of obtaining 4 tails.(Â½) indicates the probability of obtaining the remaining number of head.Therefore, the probability is 5/32.

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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