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Oh, what a happy little question! Let's break it down. The probability of getting heads on a single flip is 1/2, and the probability of getting tails is also 1/2. So, the probability of getting HTTH in that specific order on four flips would be (1/2) * (1/2) * (1/2) * (1/2) = 1/16. Just remember, there are no mistakes, just happy little accidents in probability!

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It is 1/16.

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Wiki User

8y ago
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Q: What is the P(HTTH) on four consecutive flips of a coin?
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Tommy flips a coin four times How many ways can he get three heads and one tail?

There are 4 ways to get 3 heads and 1 tail for 4 coin flips. They are: THHH, HTHH, HHTH & HHHT.


What would the theoretical probability be of flipping a coin four times?

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.


What is the probability of exactly three heads in four flips of a coin given at least two are heads?

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%.


What is the PH H H T on four flips of a coin?

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.


What is the probability of obtaining exactly four tails in five flips of a coin if at least three are tails?

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)