With one toss of a coin, there can be at most 1 head. So the probability of 4 or more heads is very definitely 0.
well, it will have 6 times of the greater chance.
50%
Let's call one coin A and the other B. omes The possible outcomes for the coins are; A heads and B tails, A tails and B heads, A and B heads, A and B tails. That's four outcomes. The possible outcomes for a single die (as in dice) are six since a die has six faces, So four times six is twenty four possible outcomes.
The answer depends on "the outcome is heads HOW MANY TIMES!"
P(Heads) = 2/3
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
The number of sequences is 27 or 128.
Coins do not have numbers, there is only the probability of heads or tails.
well, it will have 6 times of the greater chance.
Eight.
The odds are 50/50. A tossed coin does not have a memory.
mean = 5, variance = 5
The number of times a coin is tossed does not alter the probability of getting heads, which is 50% in every case, as long as the coin has not been rigged (i.e., a double-headed coin, a weighted coin) to alter the result.
It is 0.5
50%
It means that if the coin were tossed an infinite number of times, half of the tosses would come up heads and half would be tails.
Let's call one coin A and the other B. omes The possible outcomes for the coins are; A heads and B tails, A tails and B heads, A and B heads, A and B tails. That's four outcomes. The possible outcomes for a single die (as in dice) are six since a die has six faces, So four times six is twenty four possible outcomes.