The answer depends on what the experiment is!
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.
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.
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)
The answer depends on the experiment: how many coins are tossed, how often, how many dice are rolled, how often.
this isn giong to be my answerP(tails and 5) = 1 P(tails or 1) = 2
this dick
The answer to what I think the question might be, is (1/2)*(1/6) = 1/12
The answer depends on what the experiment is!
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.
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.
It is 0.5
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.
It is approx 0.2461
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.
The probability of getting five tails in a row is 1/2^5, or 1 in 32.The probability of getting five heads in a row is 1/2^5, or 1 in 32.Thus, the probability of getting either five heads or five tails in five tosses is 1 in 16.(The caret symbol means "to the power of," as in 2^5 means "2 to the 5th power.")
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)