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.
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%.
When flipping two coins, each coin has two possible outcomes: heads (H) or tails (T). The total number of outcomes when flipping two coins is 2 × 2 = 4, which includes HH, HT, TH, and TT. Out of these four outcomes, only one results in both coins landing on tails (TT). Therefore, the probability of both coins landing on tails is 1 out of 4, or 25%.
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.