Question

probability question:
Three digits are chosen in order from 0,1,2,...9.
Find the probability the digits are drawn in creasing order, if draw are made with replacement.

The answer is 0.12...but how to get that answer?

Answer 1

> creasing order

Please "iron" this out to increasing order or decreasing order. Which is it? :)

Answer 2

sorry.. it is increasing order

Answer 3

Damn. I got a different answer... still working on this....

Answer 4

Are you sure the answer is 12%? I am getting 18%.

Answer 5

By the way, this is how I got this:
If you have to choose two elements in increasing order, then if you choose 0 first, there are 9 ways to then choose larger. If you choose 1 first, there are 8, etc. On average, there are 4.5 ways. This generalizes. If you choose a number and have n numbers larger, there are on average n/2 ways to do that. I thus got (4+3.5+3+2.5+2+1.5+1+.5) % = 18%. Hope that helped ;/

Answer 6

yes I am pretty sure that the answer is 0.12

Answer 7

ummmm I don't think taking the average is viable in this kind of problems.

Answer 8

1no is the lowest so 1/9 and 2/8 and 3/7=1/4*3*7=1/74

Answer 9

1/74 is the answer

Answer 10

with replacement cannot be possible it affects

Answer 11

sorry 1/10*2/9*3/8=1/120 is the answer

Answer 12

Combination will give sorted numbers:\[\left(\begin{matrix}10 \\ 3\end{matrix}\right)=\frac{ 10·9·8 }{ 3·2·1 }=120\] And the total number of groups you can make (replacement means repetition)\[10^3=1000\]Then you have the probability is:\[P=120/1000=0.12\]