Question

(a) (8%) Compute the probability of an even integer among the 100 integers 1!, 2!, 3!, .., until 100! (here n! is n factorial or n*(n-1)*(n-2) *… 1) (b) (16%) Compute the probability of an even integer among the 100 integers: 1, 1+2, 1+2+3, 1+2+3+4, …., 1+2+3+… + 99, and 1+2+3+… + 100

Answer 1

Answer:

(a)  99%

(b)  50%

Step-by-step explanation:

(a) All factorials after 1! have 2 as a factor, so are even. Thus 99 of the 100 factorials are even, for a probability of 99%.

__

(b) The first two sums are odd; the next two sums are even. The pattern repeats every four sums. There are 25 repeats of that pattern in 100 sums, so 2/4 = 50% of sums are even.