1/2
It depends on how many points there are that the spinner can land on. If there are 8, for example, the probability would be 8/16, or 1/2...
1 out of 6
The probability of flipping a coin and having it land heads in a single flip is 1/2. To find the probability of getting heads in 6 consecutive flips, you multiply the probabilities of each individual flip: (1/2)^6. This results in a probability of 1/64, or approximately 0.0156 (1.56%).
If it a fair die, the probability is 1/3.
1/16
Odds against A = Probabillity against A / Probability for A Odds against A = (1 - Probabillity for A) / Probability for A 9.8 = (1 - Probabillity for A) / Probability for A 9.8 * Probability for A = 1 - Probability for A 10.8 * Probability for A = 1 Probability for A = 1 / 10.8 Probability for A = 0.0926
This is easiest to solve by working out the probability that no heads show and subtracting this from 1 to give the probability that at least one head shows: Assuming unbiased coins which won't land and stay on their edge, the probability of head = probability of tail = ½ → probability no heads = probability 5 tails = ½^5 = 1/32 → probability of at least one head = 1 - 1/32 = 31/32 = 0.96875 = 96.875 % = 96 7/8 %
You cannot roll "a dice" because it is one die, many dice. If you roll an ordinary, 6 faced die, the probability that it will land on 1 is 1/6.
The probability of a fair coin to land head is 1/2. Since for 4 flips it must land heads each time, the probability of 4 heads is 1/2 * 1/2 * 1/2 * 1/2 = 1/16.
the probability is 1 out of 6
The probability that the coin will land on heads each time is 1/2. (1/2) to the tenth power is 1/1024. This is the probability that the coin will not land on heads. Subtract it from one to get the probability that it will : 1-(1/1024)There is a 1023/1024 or about 99.90234% chance that the coin will land on heads at least once.(There is a 1/1024 chance that the coin will land on heads all four times.)
To find the probability of the pointer landing on 3, you need to know the total number of equal sections on the spinner. If the spinner has ( n ) sections, and one of them is labeled 3, the probability is calculated as ( \frac{1}{n} ). For example, if there are 8 sections, the probability would be ( \frac{1}{8} ). Without knowing the total number of sections, the exact probability cannot be determined.