An ordinary (fair) coin is tossed 3 times. Outcomes are thus triples of "heads" (h) and "tails" (t) which we write hth, ttt, etc.For each outcome, let N be the random variable counting the number of tails in each outcome. For example, if the outcome is hhh, then =Nhhh0. Suppose that the random variable X is defined in terms of N as follows: =X−N−N22. The values of X are given in the table below.Outcome thh ttt hht tht hhh htt tth hthValue of X −2 −8 −2 −4 −2 −4 −4 −2Calculate the probabilities P=Xx of the probability distribution of X. First, fill in the first row with the values of X. Then fill in the appropriate probabilities in the second row.Value x of X P=Xx

A coin is tossed three times. An outcome is represented by a string of the sort HTT (meaning a head on the first toss, followed by two tails). The 8 outcomes are listed in the table below. Note that each outcome has the same probability.For each of the three events in the table, check the outcome(s) that are contained in the event. Then, in the last column, enter the probability of the event.Outcomes ProbabilityHHT HTH HTT HHH THH TTH THT TTTEvent A: Alternating head and tail (with either coming first) Event B: More heads than tails Event C: A tail on the second toss

Suppose you toss a coin three times and count the number of tails. Fill in the blanks in the following table to create a probability model for this situation. Give all probabilities as a fraction or a number between 0 and 1.Group of answer choices0 tails: 28, 1 tail: 28, 2 tails: 38, 4 tails: 180 tails: 08, 1 tail: 28, 2 tails: 28, 4 tails: 180 tails: 18, 1 tail: 38, 2 tails: 38, 3 tails: 18

A coin is tossed three times. An outcome is represented by a string of the sort HTT (meaning heads on the first toss, followed by two tails).The 8 outcomes are listed below. Assume that each outcome has the same probability.Complete the following. Write your answers as fractions.

1 pointA biased coin with probability of heads 0.750.75 is tossed three times. Let 𝑋X be a random variable that represents the number of head runs, a head run being defined as a consecutive occurrence of at least two heads. Then the probability mass function of 𝑋X is given by:𝑃(𝑋=𝑥)={0.375for 𝑥 = 00.625for 𝑥 = 1P(X=x)={ 0.3750.625​ for x = 0for x = 1​ 𝑃(𝑋=𝑥)={0.297for 𝑥 = 00.703for 𝑥 = 1P(X=x)={ 0.2970.703​ for x = 0for x = 1​ 𝑃(𝑋=𝑥)={0.016for 𝑥 = 00.140for 𝑥 = 10.422for 𝑥 = 20.422for 𝑥 = 3P(X=x)= ⎩⎨⎧​ 0.0160.1400.4220.422​ for x = 0for x = 1for x = 2for x = 3​ 𝑃(𝑋=𝑥)={0.016for 𝑥 = 00.844for 𝑥 = 10.140for 𝑥 = 2P(X=x)= ⎩⎨⎧​ 0.0160.8440.140​ for x = 0for x = 1for x = 2​

An number cube (a fair die) is rolled 3 times. For each roll, we are interested in whether the roll comes up even or odd. An outcome is represented by a string of the sort oee (meaning an odd number on the first roll, an even number on the second roll, and an even number on the third roll).For each outcome, let N be the random variable counting the number of odd rolls in each outcome. For example, if the outcome is oeo, then =Noeo2. Suppose that the random variable X is defined in terms of N as follows: =X−2N2−4N3. The values of X are given in the table below.Outcome ooo eoe eeo oeo ooe eee eoo oeeValue of X 3 −5 −5 −3 −3 −3 −3 −5Calculate the probabilities P=Xx of the probability distribution of X. First, fill in the first row with the values of X. Then fill in the appropriate probabilities in the second row.Value x of X P=Xx