If the cube is uniformly weighted there is a 1 in 6 chance of any side landing face-up
1/6
The probability is 1/16.
These are all independent events. So the probability of them all happening is the product of the probabilities of each one of them happening. The desired probability is (2/6)*(1/2)*(1/2)=1/12
(1/2)^3 = 1/8th Since the initial probability (assuming independence) of getting a head in a single toss is one half (1/2) we just cube this probability because of the number of events we are performing. So if you were to try to calculate the probability of a coin being tossed 6 times it would be one half to the 6th power which is 1/64.
The experimental probability of a number cube that lands on 5 four times in a twenty toss trial is Pexp(5) = 4/20 = 1/5 = 0.20 = 20%
1/4 if they are tossed only once.
If the "number cube" you are referring to is what we normally call a die (one of a pair of dice), the chances of tossing an 8 are nothing since the cube has only 6 sides and is numbered 1 to 6. If a "number cube" is something else, perhaps someone else can answer the question.
Coins do not have numbers, there is only the probability of heads or tails.
The probability is 50-50.
7878
The probability is 0.5The probability is 0.5The probability is 0.5The probability is 0.5
50%