If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n 1)(n 2) will be divisible by 8

If n is even, then n = 2k for integers 1 ≤ k ≤ 48, and

n(n - 1)(n - 2) = 2k(2k - 1)(2k - 2) = 4k(k - 1)(2k - 1)

and this product is clearly divisible by 4. k(k - 1) is the product of two consecutive integers, and one of these must be even, so it follows that n(n - 1)(n - 2) is always divisible by 8. So

Pr[n(n - 1)(n - 2) is divisible by 8 AND n is even]

= Pr[n(n - 1)(n - 2) is divisible by 8 || n is even] * Pr[n is even]

= 1 * 1/2

= 1/2

(where I write Pr[A || B] to denote the conditional probability of A given B)

If n is odd, then n = 2k - 1 for integers 1 ≤ k ≤ 48, and

n(n - 1)(n - 2) = (2k - 1)(2k - 2)(2k - 3) = 2(k - 1)(2k - 1)(2k - 3)

This is clearly divisible by 2. We then need to find out when this is divisible by 4. We have

(k - 1)(2k - 1)(2k - 3) = 4k^3 - 12k^2 + 11k - 3

and we have divisibility by 4 if 4 divides 11k - 3. We have

11k - 3 = 0 mod 4

==>  11k = 3 mod 4

==>  33k = 3^2 = 1 mod 4

==>  (32 + 1)k = 1 mod 4

==>  k = 1 mod 4

That is, 4 divides 11k - 3 if k = 4r + 1 for integers 0 ≤ r ≤ 11. This means there are 12 integers r (and hence 12 integers k) for which 4 divides 11k - 3, which in turn means there are 12 integers n for which n(n - 1)(n - 2) is divisible 8. So

Pr[n(n - 1)(n - 2) is divisible by 8 AND n is odd]

= Pr[n(n - 1)(n - 2) is divisible by 8 || n is odd] * Pr[n is odd]

= 12/48 * 1/2

= 1/8

The cases that n is even or odd are mutually exclusive, so

Pr[n(n - 1)(n - 2) is divisible by 8 = 1/2 + 1/8 = 5/8

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

what