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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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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)
what is the p on four flips of a coin?
It is 1/16.
What is the probability that in four coin flips you get at least 2 heads?
50%
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 obtaining exactly four heads in five flips of a coin given that at least three are heads?
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.
What are the odds in favor of getting at least one head in 3 consecutive flips of a coin?
>>> 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%.
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 PHTHT on four consecutive flips of a coin?
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
What is the probability that a coin flipped four consecutive times will always land on heads?
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
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)
What is the probability of getting at least 2 heads in three flips of a fair coin?
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.
