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Q: What is the probability of obtaining exactly four heads in five flips of a coin given that at least three are heads?

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The probability of obtaining 7 heads in eight flips of a coin is:P(7H) = 8(1/2)8 = 0.03125 = 3.1%

The requirement that one coin is a head is superfluous and does not matter. The simplified question is "what is the probability of obtaining exactly six heads in seven flips of a coin?"... There are 128 permutations (27) of seven coins, or seven flips of one coin. Of these, there are seven permutations where there are exactly six heads, i.e. where there is only one tail. The probability, then, of tossing six heads in seven coin tosses is 7 in 128, or 0.0546875.

The probability of a head in one flip is 1/2. The probability of HHHHTT is (1/2)6 = 1/64 The possible correct flips are HHHHTT, HHHTHT, HHTHHT, HTHHHT, THHHHT, HHHTTH, HHTHTH, HTHHTH, THHHTH, HHTTHH, HTHTHH, THHTHH, HTTHHH, THTHHH, TTHHHH, each with a probability of 1/64. Total probability is 15/64.

It is a variable that can take a number of different values. The probability that it takes a value in any given range is determined by a random process and the value of that probability is given by the probability distribution function.It is a variable that can take a number of different values. The probability that it takes a value in any given range is determined by a random process and the value of that probability is given by the probability distribution function.It is a variable that can take a number of different values. The probability that it takes a value in any given range is determined by a random process and the value of that probability is given by the probability distribution function.It is a variable that can take a number of different values. The probability that it takes a value in any given range is determined by a random process and the value of that probability is given by the probability distribution function.

all probabilities smaller than the given probability ("at most") all probabilities larger than the given probability ("at least")

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The probability of obtaining 7 heads in eight flips of a coin is:P(7H) = 8(1/2)8 = 0.03125 = 3.1%

The requirement that one coin is a head is superfluous and does not matter. The simplified question is "what is the probability of obtaining exactly six heads in seven flips of a coin?"... There are 128 permutations (27) of seven coins, or seven flips of one coin. Of these, there are seven permutations where there are exactly six heads, i.e. where there is only one tail. The probability, then, of tossing six heads in seven coin tosses is 7 in 128, or 0.0546875.

The probability of obtaining exactly two heads in three flips of a coin is 0.5x0.5x0.5 (for the probabilities) x3 (for the number of ways it could happen). This is 0.375. However, we are told that at least one is a head, so the probability that we got 3 tails was impossible. This probability is 0.53 or 0.125. To deduct this we need to divide the probability we have by 1-0.125 0.375/(1-0.125) = approximately 0.4286

Pr(3H given >= 2H) = Pr(3H and >= 2H)/Pr(>=2H) = Pr(3H)/Pr(>=2H) = (1/4)/(11/16) = 4/11.

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%.

Assume the given event depicts flipping a fair coin and rolling a fair die. The probability of obtaining a tail is ½, and the probability of obtaining a 3 in a die is 1/6. Then, the probability of encountering these events is (½)(1/6) = 1/12.

50-50. each toss is independent of any previous toss. if all tosses are to be heads/tails then each toss you multiply by the number of chances. i,e. 2, starting with 1. three heads in a row is 1x2x2

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%

The probability that you will toss five heads in six coin tosses given that at least one is a head is the same as the probability of tossing four heads in five coin tosses1. There are 32 permutations of five coins. Five of them have four heads2. This is a probability of 5 in 32, or 0.15625. ----------------------------------------------------------------------------------- 1Simplify the problem. It asked about five heads but said that at least one was a head. That is redundant, and can be ignored. 2This problem was solved by simple inspection. If there are four heads in five coins, this means that there is one tail in five coins. That fact simplifies the calculation to five permutations exactly.

you toss 3 coins what is the probability that you get exactly 2 heads given that you get at least one head?

We have no way of knowing the probability of any given person flipping any given coin at any given time. But for any two flips of an honest coin, the probability that both are tails is 25% . (1/4, or 3 to 1 against)

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)

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