Top 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 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.

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0The probability to tossing a coin and obtaining tails is 0.5. Rolling a die has nothing to do with this outcome - it is unrelated.

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

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 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

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.

Assume the given event depicts flipping a fair coin and rolling a fair die. The probability of obtaining a tail is ½, and the probability of obtaining a 3 in a die is 1/6. Then, the probability of encountering these events is (½)(1/6) = 1/12.

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

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

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

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

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

Each flip can come up in either one of 2 ways. So the number of different historiesof three flips is 2 x 2 x 2 = 8 .One tails can come up in 3 ways in 3 flips: (H H T, H T H, or T H H).Probability of one tails is 3/8 = 37.5% .Two tails can happen in 3 ways in 3 flips: (H T T, T T H, or T H T).Probability of two tails is 3/8 = 37.5% .Three tails can only happen one way in 3 flips: T T T.Probability of three tails = 1/8 = 12.5% .No tails can only happen one way in 3 flips: H H H.Probability of no tails = 1/8 = 12.5% .The probability of any tails (1 or 2 or 3) is 7/8 = 87.5% .The probability of 2 or 3 tails is 4/8 = 50% .The probability of 1 or 3 tails is 4/8 = 50% .The probability of no tails or any tails is 8/8 = 100% .

The probability of obtaining 4 tails when a coin is flipped 4 times is: P(4T) = (1/2)4 = 1/16 = 0.0625 Then, the probability of obtaining at least 1 head when a coin is flipped 4 times is: P(at least 1 head) = 1 - 1/16 = 15/16 = 0.9375

We need to determine the separate event. Let A = obtaining four tails in five flips of coin Let B = obtaining at least three tails in five flips of coin Apply Binomial Theorem for this problem, and we have: P(A | B) = P(A ∩ B) / P(B) P(A | B) means the probability of "given event B, or if event B occurs, then event A occurs." P(A ∩ B) means the probability in which both event B and event A occur at a same time. P(B) means the probability of event B occurs. Work out each term... P(B) = (5 choose 3)(½)³(½)² + (5 choose 4)(½)4(½) + (5 choose 5)(½)5(½)0 It's obvious that P(A ∩ B) = (5 choose 4)(½)4(½) since A ∩ B represents events A and B occurring at the same time, so there must be four tails occurring in five flips of coin. Hence, you should get: P(A | B) = P(A ∩ B) / P(B) = ((5 choose 4)(½)4(½))/((5 choose 3)(½)³(½)² + (5 choose 4)(½)4(½) + (5 choose 5)(½)5(½)0)

Each time you flip a coin, the probability of getting either heads or tails is 50%.

The probability of two tails on two tosses of a coin is 0.52, or 0.25.

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.

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 of getting two tails when tossing a coin is zero, because the coin can only have one result. If, one the other hand, you toss the coin twice, then the probability of getting two tails is 0.25, i.e. the probability of one tail, 0.5, squared.

Since it is a certainty that a coin must land on either heads or tails, the probability must be 1.

The probability of tossing a coin twice and getting tails both times is 1 in 4, or 25%. If you have already tossed a coin and had it land on tails, the probability that it will land on tails again the next time you toss it is 50%.

For a normal coin, it is 0.5.

The probability is 0%. The result will be heads or it will be tails but it cannot be heads and tails.

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