Q: What is the probability of 2 coins having H and T?

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H H T T H T T H 1 in 2 chances

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 *

There are 8 permutations of three coins. Of those, three of them have only one head, so the probability of throwing only one head is 3 in 8, or 0.375. However, you said, "at most", and that includes the permutation that does not include any heads at all, so the probability of throwing at most one head in three coins 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 probability is that the remaining two outcomes are 1 H and 1 T. The probability of that is 2/4 or 50%The probability is that the remaining two outcomes are 1 H and 1 T. The probability of that is 2/4 or 50%The probability is that the remaining two outcomes are 1 H and 1 T. The probability of that is 2/4 or 50%The probability is that the remaining two outcomes are 1 H and 1 T. The probability of that is 2/4 or 50%

There are 8 permutations of three coins ... T T T T T H T H T T H H H T T H T H H H T H H H ... counting heads and sorting by count, you get ... 0 - T T T 1 - T T H 1 - T H T 1 - H T T 2 - T H H 2 - H T H 2 - H H T 3 - H H H ... so, the probability of each possible number of heads is 0: 1 in 8, 1: 3 in 8, 2: 3 in 8, and 3: 1 in 8.

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There are four different ways that two coins can land: T T T H H T H H. Only one of them is two heads, so if the coins are fair, then the probability is 1/4 = 25% .

H H T T H T T H 1 in 2 chances

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 *

When tossing a coin, there are 2 distinct possibilities: heads or tails. When tossing 4 coins, there are 2 x 2 x 2 x 2 = 16 possible outcomes. To determine the number of outcomes that will result in 2 heads and 2 tails the formula would be n!/(h!)(n-h)! where n = number of coinsand h = number of heads. In this case the number of different ways you can get 2 heads and 2 tails would be4!/(2! x 2!) = 6 Let's verify the number of different ways you can get two heads and two tails: H H T TH T T HH T H T T T H HT H H TT H T H These are the six different ways to get two heads and two tails with four coins. To find the probability, you divide 6 by the total number of possible outcomes (16) and you would get 6/16 = 3/8. The probability of tossing 4 coins and getting two heads and two tails is 3/8 or 0.375

There are 8 permutations of three coins. Of those, three of them have only one head, so the probability of throwing only one head is 3 in 8, or 0.375. However, you said, "at most", and that includes the permutation that does not include any heads at all, so the probability of throwing at most one head in three coins 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 probability is that the remaining two outcomes are 1 H and 1 T. The probability of that is 2/4 or 50%The probability is that the remaining two outcomes are 1 H and 1 T. The probability of that is 2/4 or 50%The probability is that the remaining two outcomes are 1 H and 1 T. The probability of that is 2/4 or 50%The probability is that the remaining two outcomes are 1 H and 1 T. The probability of that is 2/4 or 50%

There are 8 permutations of three coins ... T T T T T H T H T T H H H T T H T H H H T H H H ... counting heads and sorting by count, you get ... 0 - T T T 1 - T T H 1 - T H T 1 - H T T 2 - T H H 2 - H T H 2 - H H T 3 - H H H ... so, the probability of each possible number of heads is 0: 1 in 8, 1: 3 in 8, 2: 3 in 8, and 3: 1 in 8.

The answer depends on what a winner is: 1 H?, a run of 3 H?If the winner is one H, the probability of getting exactly one winner - no more no fewer - is 5/32.

The "or" in the statement is addition; so even with one coin the probability of getting a head or a tail when spinning it is 1, and since the probability can never be greater than 1, the answer is 1. Correct. I suspect the questioner means something different. Probability of getting H & H = 1/2x1/2 = 1/4 Probability of getting H & T = 1/2 x 1/2 = 1/4 Probability of getting T & H = 1/2 x 1/2 = 1/4 Probability of getting T & T = 1/2 x 1/2 = 1/4 So probability of getting just 1 head = 1/4 +1/4 = 1/2 Probability of getting 2 heads = 1/4 So probability of getting 1 or 2 heads = 1/2 + 1/4 = 3/4 Probability of not getting a head = 1/4 Similarly for switching heads and tails.

If it were the 1st one being tails and the 2nd one being heads, then you multiply the probability of each. This is 1/2 times 1/2, which is 1/4; however, since it doesn't specify first and second, that means there is also the favorable outcome of the 1st one being heads and the 2nd one tails, which is another 1/4, for a total of 1/2, or 50%. This makes sense, since all the possible outcomes are H-H, H-T, T-H, and T-T, so having a heads and a tails is 2/4, or 1/2, or 50%.

Do a binomial expansion of (T + H)400. Evaluate the 221st term for H= 1/2 and T = 1/2 and you have the answer. There is some short cut to get that term quickly, but I've forgotten it.

Tossing two coins doesn't have a probability, but the events or outcomes of tossing two coins is easy to calculate. Calling the outcomes head (H)or tails (T), the set of outcomes is: HH, HT, TH and TT as follows: 2 heads = (1/2) * (1/2) = 1/4 1 head and 1 tail, can be heads on first coin tails on second, or just the opposite, there's two possible events: (1/2)*(1/2) + (1/2)*(1/2) = 1/2 2 tails = same probability as two heads = 1/4