When a die is rolled once, the probability of a 4 showing up is 1/6.
Apply the binomial probability for finding the probability of exactly three fours out of 12 throws of a die.
n=12 (number of throws)
p=1/6 (probability of a four in a single throw)
x = 3 (number of times out of 12 , a four showing up)
P(x=3) = 12C3 (1/6)^3 (5/6)^(12-3)
= 12C3 (1/6)^3 (5/6)^9 = 0.197443
3/18 I think
It is approx 0.1974
Assuming the die is a standard die with a different number from the set {1, 2, 3, 4, 5, 6} on each side, then: Probability_of_success = number_of_ways_of_success/total_number_of_outcomes There is only 1 way to roll a 4 and there are 6 possible outcomes, therefore the probability of rolling a 4 is 1/6 The die has no knowledge of previous rolls; each roll is independent, thus: The probability of three fours is a row is the probability of a 4 times the probability of a 4 times the probability of a 4, which is: Probability = 1/6 × 1/6 × 1/6 = 1/216
The probability of 3 specific dice rolls is the probability that each one will happen multiplied together. For instance, the probability of rolling 2 then 6 then 4 is the probability of all of these multiplied together: The probability of rolling 2 is 1/6. The probability of rolling 6 is 1/6. The probability of rolling 4 is 1/6. Multiply these together and we get the total probability as 1/216
Anywhere from 0 to 1; it depends on the shape and what numbers are written on the faces.
3/18 I think
The probability of getting three fives in the first three rolls and non-fives in the next three rolls is; P(5,5,5,N5,N5,N5) = 1/6 x 1/6 x 1/6 x 5/6 x 5/6 x 5/6 = 0.002679... The number of different order in which the fives can come out is given by; 6C3 = 6!/[3!∙(6-3)!] = 20 So the probability that in 6 rolls of a fair die exactly three fives (in any order) will come out is; P(three fives any order) = (20)∙(1/6)3∙(5/6)3 = 0.05358... ~ 5.4%
With a fair die, it is 1/216 in three rolls, but the probability increases to 1 (a certainty) as the number of rolls is increased.
It is approx 0.1974
Assuming the die is a standard die with a different number from the set {1, 2, 3, 4, 5, 6} on each side, then: Probability_of_success = number_of_ways_of_success/total_number_of_outcomes There is only 1 way to roll a 4 and there are 6 possible outcomes, therefore the probability of rolling a 4 is 1/6 The die has no knowledge of previous rolls; each roll is independent, thus: The probability of three fours is a row is the probability of a 4 times the probability of a 4 times the probability of a 4, which is: Probability = 1/6 × 1/6 × 1/6 = 1/216
The probability of 3 specific dice rolls is the probability that each one will happen multiplied together. For instance, the probability of rolling 2 then 6 then 4 is the probability of all of these multiplied together: The probability of rolling 2 is 1/6. The probability of rolling 6 is 1/6. The probability of rolling 4 is 1/6. Multiply these together and we get the total probability as 1/216
The probability is 90/216 = 5/12
To find the likelihood that Tammy wins by rolling at least one three in her first four rolls of a standard die, we can calculate the probability of losing (not rolling a three) and subtract it from 1. The probability of not rolling a three on a single roll is ( \frac{5}{6} ). Therefore, the probability of not rolling a three in four rolls is ( \left(\frac{5}{6}\right)^4 \approx 0.4823 ). Thus, the probability that Tammy wins by rolling at least one three in four rolls is ( 1 - \left(\frac{5}{6}\right)^4 \approx 0.5177 ), or about 51.77%.
Anywhere from 0 to 1; it depends on the shape and what numbers are written on the faces.
33%
To find the overall probability, we first look at the outcomes of rolling a die. There are three even numbers (2, 4, 6) and three odd numbers (1, 3, 5). If William rolls an even number (probability of 3/6 or 1/2), he flips a coin, which has a probability of 1/2 for heads or tails. If he rolls an odd number (also 1/2), he rolls again, and this process continues. The final probability of outcomes depends on the specific event of interest (e.g., getting heads after flipping the coin), so more context is needed to provide a specific probability.
Rolling a sum of 15 on three rolls of a die, when the first roll is a 4, is the same as rolling a sum of 11 on the second and third roll. The probability of rolling 11 on two dice is 3 in 36, or 1 in 12.