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Q: How many possible outcomes if each coin is flipped once?

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There are two outcomes for each coin and three coins; 2 x 2 x 2 = 23 = 8 outcomes.

Two possible outcomes for each flip. 2,048 possible histories of 11 flips.

In three flips of a fair coin, there are a total of 8 possible outcomes: T, T, T; T, T, H; T, H, T; T, H, H; H, H, H; H, H, T; H, T, H; H, T, T Of the possible outcomes, four of them (half) contain at least two heads, as can be seen by inspection. Note: In flipping a coin, there are two possible outcomes at each flipping event. The number of possible outcomes expands as a function of the number of times the coin is flipped. One flip, two possible outcomes. Two flips, four possible outcomes. Three flips, eight possible outcomes. Four flips, sixteen possible outcomes. It appears that the number of possible outcomes is a power of the number of possible outcomes, which is two. 21 = 2, 22 = 4, 23 = 8, 24 = 16, .... Looks like a pattern developing there. Welcome to this variant of permutations.

7/8

If each coin is a different color, then there are 32 possible outcomes. If you can't tell the difference between the coins, and you're just counting the number of heads and tails, then there are 6 possible outcomes: 5 heads 4 heads 3 heads 2 heads 1 heads all tails

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There are two outcomes for each coin and three coins; 2 x 2 x 2 = 23 = 8 outcomes.

If you can identify the outcomes with who flipped each coin: eg Joe and Mary = Heads, Sam = Tails, then 23 = 8. Otherwise, 4.

Two possible outcomes for each flip. 2,048 possible histories of 11 flips.

Heads or tails; each have a probability of 0.5 (assuming a fair coin).

Each coin can land in two ways.The die has 6 possible outcomes.So there are 2 x 2 x 6 = 24 possible outcomes for the whole experiment.Note that I am assuming the coins can be told apart - say the first coin and 2nd coin and that H and then T is different that T and then H. If not, then there are only be three outcomes for the coins-- 2 heads, 1 head or no heads and the total number of outcomes would be 3 X 6 = 18.

In three flips of a fair coin, there are a total of 8 possible outcomes: T, T, T; T, T, H; T, H, T; T, H, H; H, H, H; H, H, T; H, T, H; H, T, T Of the possible outcomes, four of them (half) contain at least two heads, as can be seen by inspection. Note: In flipping a coin, there are two possible outcomes at each flipping event. The number of possible outcomes expands as a function of the number of times the coin is flipped. One flip, two possible outcomes. Two flips, four possible outcomes. Three flips, eight possible outcomes. Four flips, sixteen possible outcomes. It appears that the number of possible outcomes is a power of the number of possible outcomes, which is two. 21 = 2, 22 = 4, 23 = 8, 24 = 16, .... Looks like a pattern developing there. Welcome to this variant of permutations.

There are technically 8 possible outcomes if you are talking about the side of the coin it lands on. Each coin has 2 possible outcomes (landing on heads and landing on tails). To figure out the number of outcomes for all the coins you multiply the outcomes for all of the coins together: 2 X 2 X 2= 8.

I am guessing SamJoe, means SAM and JOE not one person, so three people flip a coin, we have two outcomes each times, so 23= 8 possible outcomes. If you had n people, there would be 2n outcomes. For example, if two people flip there are 4 outcomes HH TT HT or TH

Each toss has 2 outcomes; so the number of outcomes for 3 tosses is 2*2*2 = 8

It is not. There are only two possible outcomes for each toss of a coin whereas the number of possible outcomes when selecting a marble from a bag will depend on the numbers of distinct marbles in each bag. The coin toss generates a binomial distribution the marbles experiment is multinomial.

There are 36 possible outcomes: 6 for each die.

Each coin has two possible outcomes, either Heads or Tails. Then the number of outcomes when all 4 coins are tossed is, 2 x 2 x 2 x 2 = 16.

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