If you roll a single die and count the number of dots on top, what is the sample space of all possible outcomes? are the outcomes equally likely? 0, 1, 2, 3, 4, 5, 6; not equally likely 0, 1, 2, 3, 4, 5, 6; equally likely 1, 2, 3, 4, 5, 6; not equally likely 1, 2, 3, 4, 5, 6; equally likely (b) assign probabilities to the outcomes of the sample space of part (a). (enter your answers as fractions.) outcome probability 1 2 3 4 5 6 do the probabilities add up to 1? should they add up to 1? explain. yes, but they should not because these values do not cover the entire sample space. yes, because these values cover the entire sample space. no, because these values do not cover the entire sample space. no, but they should because these values cover the entire sample space. (c) what is the probability of getting a number less than 6 on a single throw? (enter your answer as a fraction.) (d) what is the probability of getting 3 or 4 on a single throw? (enter your answer as a fraction.)

(a) the sample space is {1, 2, 3, 4, 5, 6}. The outcomes 0, 1, 2, 3, 4, 5, 6 are not equally likely. The outcomes 1, 2, 3, 4, 5, 6 are equally likely.

(b) If X is the random variable who shows the number of dots on top of the die, then, P(X = 1) = P(X = 2) = P(X = 3) = P(X = 4) = P(X = 5) = P(X = 6) = 1/6.

Yes, outcome probabilities should add up to 1 because the values 1, 2, 3, 4, 5, 6 cover the entire sample space.

(c) P(X < 6) = 5/6

(d) P(X = 3 o X = 4) = 1/3

Step-by-step explanation:

Let X be the random variable who shows the number of dots on top of the die

(a) A single die has six sides, the number of dots in each side is 1, 2, 3, 4, 5 or 6, in a throw of a sigle die, we can get any of these numbers, therefore, the sample space is {1, 2, 3, 4, 5, 6}

The outcomes 0, 1, 2, 3, 4, 5, 6 are not equally likely because the probability of getting a 0 is zero. There is not number 0 in some side of a die.

The outcomes 1, 2, 3, 4, 5, 6 are equally likely. The probability of getting any side when throwing a die is the same.

(b)  P(X = 1) = P(X = 2) = P(X = 3) = P(X = 4) = P(X = 5) = P(X = 6) = 1/6 because the probability of getting any side of a die is the same. The probabilities should add up to 1 because 1, 2, 3, 4, 5, 6 cover the entire sample space.

(c) The probability of getting a number less than 6 on a single throw is

P(X<6) = P[(X=1) ∪(X=2) ∪ (X=3) ∪(X=4) ∪(X=5)] = P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5) (because the events are mutually exclusive) = 1/6 +1/6+1/6+1/6+1/6=5/6

(d) P(X=3 o X=4) = P[(X=3)∪(X=4)]=P(X=3)+P(X=4) (because the events are mutually exclusive) = 1/6 + 1/6 = 2/6 = 1/3

the balance is found at the link

step-by-step explanation: