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2010-11-09 21:33:49
2010-11-09 21:33:49

There are 210 total possible outcomes from flipping a coin 10 times.

There is one possible outcome where there are 0 heads.

There are 10 possible outcomes where there is 1 head.

So there are 210 - 11 possible outcomes with at least 2 heads.



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Number of possible outcomes of one coin = 2Number of possible outcomes of six coins = 2 x 2 x 2 x 2 x 2 x 2 = 64Number of possible outcomes with six heads = 1Probability of six heads = 1/64Probability of not six heads = at least one tails = 63/64 = 98.4375%

There are 2^5 (2*2*2*2*2), or 32, possible outcomes of tossing a coin 5 times. Only one of those outcomes does not contain any tails. This leaves us with 31/32, or 97% chance of at least one toss coming up tails.

The probability of tossing heads on all of the first six tosses of a fair coin is 0.56, or 0.015625. The probability of tossing heads on at least one of the first six tosses of a fair coin is 1 - 0.56, or 0.984375.

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%.

In three flips of a fair coin, there are a total of 8 possible outcomes: T, T, T; T, T, H; T, H, T; T, H, H; H, H, H; H, H, T; H, T, H; H, T, T Of the possible outcomes, four of them (half) contain at least two heads, as can be seen by inspection. Note: In flipping a coin, there are two possible outcomes at each flipping event. The number of possible outcomes expands as a function of the number of times the coin is flipped. One flip, two possible outcomes. Two flips, four possible outcomes. Three flips, eight possible outcomes. Four flips, sixteen possible outcomes. It appears that the number of possible outcomes is a power of the number of possible outcomes, which is two. 21 = 2, 22 = 4, 23 = 8, 24 = 16, .... Looks like a pattern developing there. Welcome to this variant of permutations.

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

There are four outcomes possible (not considering order)HHHHHTHTTTTTOnly in two of the cases are there two or more headsThe probability is 0.5

The probability of tossing a coin 9 times and getting at least one tail is: P(9 times, at least 1 tail) = 1 - P(9 heads) = 1 - (0.50)9 = 0.9980... ≈ 99.8%

The answer is not easy: it depends on the situation that you are studying. It is simple to identify all the possible outcomes of rolling a die and tossing a coin: it is the set of ordered pairs defined as the set {(n, a), where n is an integer from 1 to 6 and a is H or T}. In general, it is not at all simple to determine all possible outcomes. There may be outcomes that you did not expect or and make allowance for. In practise, therefore, it is sometimes best to have a category for "other outcomes". In principal, a tossed coin could end up on its edge, or at least, end up leaning against some obstruction so that it is not flat on H or T.

By tossing two coins the possible outcomes are:H & HH & TT & HT & TThus the probability of getting exactly 1 head is 2 out 4 or 50%. If the question was what is the probability of getting at least 1 head then the probability is 3 out of 4 or 75%

The probability of tossing 6 heads in 6 dice is 1 in 26, or 1 in 64, or 0.015625. THe probability of doing that at least once in six trials, then, is 6 in 26, or 6 in 64, or 3 in 32, or 0.09375.

There are 8 permutations of three coins. Of these, 3 of them have two heads, so the probability of tossing two heads on three coins is 3 in 8, or 0.375. However, you said, "at least", so that includes the case of three heads, so the probability of throwing at least two heads is 4 in 8, or 0.5. T T T T T H T H T T H H * H T T H T H * H H T * H H H *

The possible outcomes are: hhh hht hth thh htt tht tth ttt If you mean exactly 2 heads (and not 3 heads), then you can see 3 out of 8 possibilities have exactly 2 heads. So the probability is 3/8 or 0.375 If you mean at least 2 heads (2 or 3 heads), then 4/8 or 0.5.

9/16Answer:There are 16 possible outcomes. The criteria (at least two heads) are met if the coins come up in any one of 11 different ways:One way with four heads: HHHHFour ways with three heads: HHHT, HHTH, HTHH, THHHSix ways with two heads: HHTT, HTHT, THHT, THTH, TTHH, HTTHThis would make the probability 11/16

There are 23 or 8 possible outcomes listed below (H=Head, T=Tails). HHH HHT HTH HTT THH THT TTH TTT each has a 1/8 probability. Count the number that has at least 2 heads and it is 3. So, the probability is 3/8.

Okay, lets write out the possible outcomes when flipping a coin 3 times: HHH, HHT, HTH, THH, TTH,THT,HTT,TTT That constitures 8 scenarios in which the coin can fall over a 3 flip trial. Now, it is known that you got "at least one head" so therefore we can rule out the no head scenario (TTT) which leaves us with 7. Of those 7 times, how many times does it fall heads exactly twice? Well, we have HHT,HTH,THH. From this you can say that it there are 3 possible outcomes in which you get exactly two heads given that you get at least one head. 3/7.

No matter how many coins are thrown, the possibility of having AT LEAST ONE 'head' is 50%. This changes if you specify the number of 'heads' that must be shown.

3 coins can land in 8 different ways. Only one of these ways is all tails. So the probability of rolling at least one heads is 7/8 = 87.5% .

There are 36 possible combinations. Eleven of them have at least one four in it. That means it is 11 over 36, which is a 30.55% chance.

The total number of outcomes you could get by flipping a coin 4 times is 2^4 or 16 ways as each coin toss yields two possible outcomes (Heads or Tails) and there are four trials. With that said, you need to find out how many ways there are to get 3 heads or 4 heads. You could use combinations to find this: n C r = n! / [ r! (n-r)! ] 4 C 3 = 4! / [ 3! (4-3)! ] 4 C 3 = 4! / 3! 4 C 3 = 4 4 C 4 = 4! / [ 4! (4-4)! ] 4 C 4 = 4! / 4! 4 C 4 = 1 The total numbers of ways that one could get at least 3 H heads is equal to five (4 ways to get three heads, and one way to get 4 heads). So the probability of getting at least 3 heads in four tosses is equal to 5/16 or 31.25%.

No, at least not at this point in the evolution of medical science.

If you mean 'at least' 2 heads, the probability is 50%. If you mean exactly 2, the probability is 3/8, or 37.5%. There are 3 independent coin tosses, each of which is equally likely to come up heads or tails. That's a total of 2 * 2 * 2 or 8 possible outcomes (HHH, HHT, HTH, etc.). Of these, 4 include 2 or 3 heads, which is half of 8. Only 3 include exactly 2 heads, so the probability of that is 3/8.

The empirical probability can only be determined by carrying out the experiment a very large number of times. Otherwise it would be the theoretical probability.

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.

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