It is 1/16.
There are 4 ways to get 3 heads and 1 tail for 4 coin flips. They are: THHH, HTHH, HHTH & HHHT.
Anyone can flip a coin four times so I say 100 percent probability. On the other maybe you should ask the odds of the results from four flips.
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%.
Assuming the coin is fair, the probability of that sequence is 1/16. The probability of three H and one T, in any order, is 1/4.
If at least three flips are tails, there are two scenarios where we can obtain exactly four tails in five flips. Either the first four flips are tails and the last flip is heads, or the first flip is heads and the next four flips are tails. Each scenario has a probability of 1/2 * 1/2 * 1/2 * 1/2 * 1/2 = 1/32. Therefore, the probability of obtaining exactly four tails in five flips if at least three are tails is 1/32 + 1/32 = 1/16.
It is 1/16.
50%
There are 4 ways to get 3 heads and 1 tail for 4 coin flips. They are: THHH, HTHH, HHTH & HHHT.
Anyone can flip a coin four times so I say 100 percent probability. On the other maybe you should ask the odds of the results from four flips.
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.
>>> 1:7 (or, if you like probability, 87.5%)I disagree. There are four possible combinations of three tosses (where order does not matter):HHHHHTHTTTTTThree of these combinations will show at least one head - only by throwing three tails will you not throw at least one head.Thus, the probability of throwing at least one head in three flips is 75%.
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%.
Assuming the coin is fair, the probability of that sequence is 1/16. The probability of three H and one T, in any order, is 1/4.
The result of each event (or flip) is independent of the others. Therefore, the individual probabilities can be multiplied. P(HTHT) = Probability of (heads-tails-heads-tails): 1/2 x 1/2 x 1/2 x 1/2 = 1/16
The probability that a coin flipped four consecutive times will always land on heads is 1 in 16. Since the events are sequentially unrelated, take the probability of heads in 1 try, 0.5, and raise that to the power of 4... 1 in 24 = 1 in 16
If at least three flips are tails, there are two scenarios where we can obtain exactly four tails in five flips. Either the first four flips are tails and the last flip is heads, or the first flip is heads and the next four flips are tails. Each scenario has a probability of 1/2 * 1/2 * 1/2 * 1/2 * 1/2 = 1/32. Therefore, the probability of obtaining exactly four tails in five flips if at least three are tails is 1/32 + 1/32 = 1/16.
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.