Answer:
(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
