Math and Arithmetic

Statistics

Probability

123

We use three coins (quarter, nickel, dime) each are flipped only once. We get 8 possible outcomes (or four outcomes as an alternative).

enless you include it landing on it's side the two possible outcomes for this are: Heads and Tails

A system with two possible outcomes with equal probabilities.

If you know which coin is which, there are 16possible outcomes.If you're only counting the number of Heads and Tails, there are 5 .

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

The event is that the coin lands with the Heads on top.

Since each coin would have the outcome with Heads and Tails: Then among the 32 coins, we can have the possible outcomes from no Heads, 1 Head, 2 Heads, ....... , 31 Heads, 32 Heads. Therefore we would have 33 outcomes.

Each coin can come out either heads (H) or tales (T). Since you're tossing four coins at once, I'm assuming there is no sense of order to be accounted for. In that case, the possible outcomes are the following: HHHH HHHT HHTT HTTT TTTT

To determine the amount of possible outcomes, there must be a number of sections for each spinner

If a normal die is rolled once, the outcomes are {1, 2, 3, 4, 5, 6}.

The answer depends on what the outcome comprises.If the outcome is simply a count of H and T, then there are 12 outcomes: 0H, 1H, 2H, ... 10H and 11H (with the corresponding number of T in each case).If the result for each coin is recorded individually, there are 211 = 2048 possible outcomes.There are many other results that can be defined, and the number of possible outcomes will vary according to the definitions.The answer depends on what the outcome comprises.If the outcome is simply a count of H and T, then there are 12 outcomes: 0H, 1H, 2H, ... 10H and 11H (with the corresponding number of T in each case).If the result for each coin is recorded individually, there are 211 = 2048 possible outcomes.There are many other results that can be defined, and the number of possible outcomes will vary according to the definitions.The answer depends on what the outcome comprises.If the outcome is simply a count of H and T, then there are 12 outcomes: 0H, 1H, 2H, ... 10H and 11H (with the corresponding number of T in each case).If the result for each coin is recorded individually, there are 211 = 2048 possible outcomes.There are many other results that can be defined, and the number of possible outcomes will vary according to the definitions.The answer depends on what the outcome comprises.If the outcome is simply a count of H and T, then there are 12 outcomes: 0H, 1H, 2H, ... 10H and 11H (with the corresponding number of T in each case).If the result for each coin is recorded individually, there are 211 = 2048 possible outcomes.There are many other results that can be defined, and the number of possible outcomes will vary according to the definitions.

Number of possible outcomes of one coin = 2Number of possible outcomes of six coins = 2 x 2 x 2 x 2 x 2 x 2 = 64Number of possible outcomes with six heads = 1Probability of six heads = 1/64Probability of not six heads = at least one tails = 63/64 = 98.4375%

Your question is a bit difficult to understand. I will rephrase it as follows: What is the probability of getting a head if a coin is flipped once? p = 0.5 What is the probability of getting 2 heads if a coin is flipped twice = The possible events are HT, TH, HH, TT amd all are equally likely. So the probability of HH is 0.25. What is the probability of getting at least on head if the coin is flipped twice. Of the possible events listed above, HT, TH and HH would satisfy the condition of one or more heads, so the probability is 3 x 0.25 = 0.75 or 3/4. Also, since the probability of TT is 0.25, and the probability of all events must sum to 1, then we calculate the probability of one or more heads to be 1-0.25 = 0.75

Assuming it is a 6-sided number cube, it would be 6.

Since there are 6 possible outcomes, and you want the probability of obtaining one of the outcomes (in your case 6), the probability of it landing on a 6 is 1/6.

18 different combinations. When a coin is tossed twice there are four possible outcomes, (H,H), (H,T), (T,H) and (T,T) considering the order in which they appear (first or second). But if we are talking of combinations of the two individual events, then the order in which they come out is not considered. So for this case the number of combinations is three: (H,H), (H,T) and (T,T). For the case of tossing a die once there are six possible events. The number of different combinations when tossing a coin twice and a die once is: 3x6 = 18 different combinations.

The probability is 25%. The probability of flipping a coin once and getting heads is 50%. In your example, you get heads twice -- over the course of 2 flips. So there are two 50% probabilities that you need to combine to get the probability for getting two heads in two flips. So turn 50% into a decimal --> 0.5 Multiply the two 50% probabilities together --> 0.5 x 0.5 = 0.25. Therefore, 0.25 or 25% is the probability of flipping a coin twice and getting heads both times.

The answer in the back of the book says that it is: TT, TH, H1, H2 H3, H4, H5, H6= 8 possibilities. But I don't understand what that means or how the answer was found. Help?

There is 24 or 16 outcomes. There is 4 ways to get heads once (HTTT & THTT & TTHT & TTTH). So, the probability of getting heads only once if a fair coin is tossed 4 times is 4/16 or 1/4.

There are 72 permutations of two dice and one coin.

1st dice = 6 possibilities 2nd dice = 6 3rd dice = 6 Coin = 2 Total = 6x6x6x2 = 432 outcomes

The sample space, with a fair coin, is {Heads, Tails}.I am assuming that the probability that the coin ends up resting on its edge is so small that it can be ignored as a possible outcome.

Ignoring the very small probability of the coins landing on their edge, the probability of header = probability of tails = 1/2 Pr(2 tails) = ₃C₂ × ½² × ½¹ = 3 × ½³ = 3/8 --------------------------------------------- This is small enough to list the possible outcomes: Outcomes = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} (In each outcome the first letter is the first coin, 2nd letter the second coin and the third letter the third coin, which can be either H = Heads, or T = Tails.) There are 8 outcomes of which {HTT, THT, TTH} 3 satisfy having 2 tails → pr(2 tails) = 3/8

It is 1/(2^4) i.e. 1/16 This is because there is 1 in 2 chance of it happening once, and multiply by itself for each successive go.

Assuming this is a standard die then this is simply a list of the numbers ie 1, 2, 3, 4, 5, 6.

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