1/6 = 16.67%
the probability is 2/6
When a number cube is rolled twice, there are 36 possible outcomes. (1,1),(1,2),....(6,6). (3,3) occurs only once. Therefore, the probability of rolling a 3 both times is 1/36.
When a die is rolled once, the probability of 1 showing up is 1/6 . When a die is rolled 14 times, the probability of 1 showing up 6 times is a binomial probability. Let x = number of times 1 shows up P(x=6) = 14C6 (1/6)^6 (5/6)^(14-6) = 14C6 (1/6)^6 (5/6)^8 = 0.014982
There are 6 sides on a die, so the denominator should be 6. The number 3 appears on the dice once, so the fraction probability should be 1/6.
if you flip a coin once, the chance it will be heads is 50%
If it is a fair die that is rolled once, then the probability is 2/3.
5/6 = 83.3%
7
It is 1/2 if it is a fair number cube which is rolled once.
it depends on how many numbers are on the number cube. if it only goes to 6, then the probability is 1/6. if it goes to 8 then the probability is 2/8. and so on and so on...
1 out of 2
The probability is 1, if the dice are rolled often enough.
There is a 1 in 2 chance, so 1/2.
The probability is 1. It is a certainty that you will roll a number between and including one and six. The probability of rolling each individual number is 1/6.
P(1 or 6)
the probability is 2/6
To find the probability of getting an odd number at least once when a die is tossed thrice, we can use the complementary approach. The probability of not getting an odd number (i.e., getting an even number) in a single toss is ( \frac{3}{6} = \frac{1}{2} ). Therefore, the probability of getting an even number in all three tosses is ( \left(\frac{1}{2}\right)^3 = \frac{1}{8} ). Thus, the probability of getting an odd number at least once is ( 1 - \frac{1}{8} = \frac{7}{8} ).