Suppose four integers are chosen successively at random between 0 and 12, inclusive. Find the probability that:

a. they are all different?

b. not more than 2 are the same?

Guest Feb 18, 2020

#2**+1 **

I will give this a try!!

a - The first integer can be any integer between 0 and 12 =13 / 13

The probability that the 2nd number is different from the first =12 / 13

The probability that the 3rd number is different from the first two =11 / 13

The probability that the 4th number is different from the first three =10 / 13

Then, the overall probability that all four numbers are different from each other =13 /13 x 12 / 13 x 11 / 13 x 10 / 13 =17,160 / 13^4 =1,320 / 2,197 =60.08 %

b - [13 C 2 x 13^11] / 13^13 =6 / 13 =46.15 %

Note: Somebody should check this,please. Thanks.

Guest Feb 18, 2020