The probabilty would be 3:18 or 1:6 because you have 3 dice all have 6 sides you do 3*6=18 for the denominator. Now for the numoratoreach die has numbers 1-6 and theres is only for on each dice and there is 3 dice so you would end up having 3 as the numorator. To get one fith you have to simplifie but i will go onto that next time.
50%. There are three dice; each die has 6 faces.
The answer depends on what the experiment is!
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
Probably 3/4
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%. There are three dice; each die has 6 faces.
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%.
The answer depends on what the experiment is!
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
Probably 3/4
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.
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.
The probability of rolling a 3 on a six-sided die in a single roll is 1/6. When rolling the die three times, the probability of getting at least one 3 can be calculated using the complement: first, find the probability of not rolling a 3 in three rolls, which is (5/6)³. Subtract this value from 1 to find the probability of rolling at least one 3 in three attempts.
Possible outcomes of one roll = 6.Probability of an even number on one roll = 3/6 = 0.5 .Probability of an even number on the second roll = 0.5 .Probability of an even number on the third roll = 0.5 .Probability of an even number on all three rolls = (0.5 x 0.5 x 0.5) = 0.125 = 1/8The probability of at least one odd number is the probability of not gettingan even number on all 3 rolls. That's (1 - 1/8) = 7/8 or 0.875 or 87.5% .
It is 0.5
It is approx 0.1974
It is (1/6)5 = 1/7776 = 0.00013 approx.