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Assume the coin is fair, so there are equal amount of probabilities for the choices.
There are two possible choices for a flip of a fair coin - either a head or a tail. The probability of getting a head is ½. Similarly, the probability of getting a tail is ½.
Use Binomial to work out this problem. You should get:
(5 choose 4)(½)4(½).
- (5 choose 4) indicates the total number of ways to obtain 4 tails in 5 flips.
- (½)4 indicates the probability of obtaining 4 tails.
- (½) indicates the probability of obtaining the remaining number of head.
Therefore, the probability is 5/32.
What else can I help you with?
What is the probability of tossing tails twice on two flips of a coin?
1 in 4
What is the expected number of flips of a coin to simulate a six-sided die?
Five coin flips. Any outcome on a six-sided die has a probability of 1 in 6. If I assume that the order of the outcome does not matter, the same probability can be achieved with five flips of the coin. The possible outcomes of five flips of a coin are as follows: 5 Heads 5 Tails 4 Heads and 1 Tails 4 Tails and 1 Heads 3 Heads and 2 Tails 3 Tails and 2 Heads For six possible outcomes.
What is the probability of obtaining exactly 5 heads in 6 flips of a fair coin?
It is approx 0.0938
What is the probability of obtaining exactly two heads in three flips of a coin given that at least one is a head?
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
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 obtaining exactly 4 tails from the 9 flips of the coin?
It is approx 0.2461
Suppose that you toss a coin and roll and die What is the probability of obtaining tails?
The probability to tossing a coin and obtaining tails is 0.5. Rolling a die has nothing to do with this outcome - it is unrelated.
What is the probability of obtaining exactly four tails in five flips?
Assume the coin is fair, so there are equal amount of probabilities for the choices.There are two possible choices for a flip of a fair coin - either a head or a tail. The probability of getting a head is ½. Similarly, the probability of getting a tail is ½.Use Binomial to work out this problem. You should get:(5 choose 4)(½)4(½).(5 choose 4) indicates the total number of ways to obtain 4 tails in 5 flips.(½)4 indicates the probability of obtaining 4 tails.(½) indicates the probability of obtaining the remaining number of head.Therefore, the probability is 5/32.
What is the probability of 3 heads and 2 tails on five flips of a coin?
It is 0.3125
What is the probability of tossing tails twice on two flips of a coin?
1 in 4
What is the probability of obtaining exactly seven heads in eight flips of a coin given that at least one is a head?
The probability of obtaining 7 heads in eight flips of a coin is:P(7H) = 8(1/2)8 = 0.03125 = 3.1%
What is the expected number of flips of a coin to simulate a six-sided die?
Five coin flips. Any outcome on a six-sided die has a probability of 1 in 6. If I assume that the order of the outcome does not matter, the same probability can be achieved with five flips of the coin. The possible outcomes of five flips of a coin are as follows: 5 Heads 5 Tails 4 Heads and 1 Tails 4 Tails and 1 Heads 3 Heads and 2 Tails 3 Tails and 2 Heads For six possible outcomes.
What is the probability of obtaining exactly 5 heads in 6 flips of a fair coin?
It is approx 0.0938
What is the probability of obtaining exactly two heads in three flips of a coin given that at least one is a head?
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
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 2 heads in a row followed by two tails on 4 flips of a coin?
1/16
What is the probability of obtaining exactly six heads in seven flips of a coin given that at least one is a head?
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.
