Give an example of three random events x, y, z for which any pair are independent but all three are not mutually independent.

We throw 2 fair dice

X = 1 if the first dice is equal to 3, 0 otherwise

Y = 1 if the second dice is equal to 3, 0 otherwise

Z = 1 if the sum of the dice in even, 0 otherwise

Step-by-step explanation:

X and Y are independent because X depends on the first dice and Y depends on the second one. Regarless of the results of the second dice, the probability of getting a 3 in the first one is 1/6 and independently of the results of the fisrt dice, the probability of getting 3 in the second toss is also 1/6.

Lets see that X and Z are independent, a similar argument can be used to show that Y and Z are independent.

If the first dice is 3, then in order for the sum to be even we need the second dice to be odd, hence the second dice has to be either 1, 3 or 5 and the probability for that is 1/2.

If the first dice is even, then the second one should be even, and if the second dice is odd, then the second dice should be odd. Since a dice being even (or odd) has a probability of 1/2, then using the theorem of total probability, we have

P(Z) = P(Z | First dice even) * P(first dice even) + P(Z | First dice odd) * P(first dice odd) = P(second dice even)(first dice even)+ P(second dice odd)*P(first dice odd) = 1/2*1/2 + 1/2*1/2 = 1/4+1/4 = 1/2.

Hence, the probability of Z doesnt change if we condition it by X.

Now, lets calculate P(X | Z)

If Z is true, then both dices are either both odd or either both even. Both cases with 1/2 probability. If the dices are even, then X has a probability 0 of happening, however if both dices are odd, then the first dice could be either 1, 3 or 5, and the probability of it being 3 is 1/3, hence

P(X| Z) = 1/2 * 0 + 1/2 * 1/3 = 1/6 = P(X)

This shows that X and Z are independent.

Then X, Y and Z are independent in pairs, however ther are not mutually independent. If we assume that X and Y are valid, then the sum of the dices is 3+3 = 6, hence Z will always happen. As a result

P ( Z | X, Y) = 1

But

P(Z) = 1/2

Hence, they cant be independent.

-12 > a

step-by-step explanation:

-11 > 1 + a

subtract 1 from each side

-11-1 > 1-1+a

-12 > a

step-by-step explanation:

i found the life tables for the year 2012. 

life table of males at age 30. probability of dying between ages x and x+1

total population:  

29–30 . . . . . . . . . . . . 0.001030

30–31 . . . . . . . . . . . . 0.001061

males:  

29–30 . . . . . . . . . . . . 0.001420

30–31 . . . . . . . . . . . . 0.001448

white population

29–30 . . . . . . . . . . . . 0.001005

30–31 . . . . . . . . . . . . 0.001037

white males

29–30 . . . . . . . . . . . . 0.001372

30–31 . . . . . . . . . . . . 0.001399

non hispanic white population

29–30 . . . . . . . . . . . . 0.001083

30–31 . . . . . . . . . . . . 0.001116

non hispanic white males

29–30 . . . . . . . . . . . . 0.001477

30–31 . . . . . . . . . . . . 0.001503

jacob, non hispanic white male. 30 years old.

probability that jacob lives to be 30 years old. 

since the given probabilities are probabilities of dying, i think we simply subtract it from 1 to get the probability of living. we will use the probability of non hispanic white male.

                                          probability of living to be 30 y.o.

29 - 30 : 1 - 0.001477 = 0.998523

30 - 31 : 1 - 0.001503 = 0.998497