# How do you do this? Probability

A computer prints out three digits chosen at random from 0 to 9 inclusive. Find the probability that:

a) all three digits are different

b) the third digit differs from the first two digits

For (a) I went $$1/_9C_3 = 1/84$$. I think it's wrong but I didn't know what else to try.

And for (b) I don't even know where to start.

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lurflurf
Homework Helper
ms. confused said:
A computer prints out three digits chosen at random from 0 to 9 inclusive. Find the probability that:

a) all three digits are different

b) the third digit differs from the first two digits

For (a) I went $$1/_9C_3 = 1/84$$. I think it's wrong but I didn't know what else to try.

And for (b) I don't even know where to start.
In both of these there are 10^3=1000 possibilities so
probability=(# of possibilities that meat criteria)/1000
A) try nPr since order matters.
2) given the two numbers they are written AAB figure how many pairs of numbers are possible.

Last edited:
The reason that your first answer is wrong that you havent used correctly the definition of probability : it is favourable cases/total possible cases:

For the first one:

Total possible numbers are 10 and not 9 , count them down if you have any confusion.
Total number of favourable cases are $10 P_3$ and total possible cases are $10^3$.Divide them to get the answer.

For the second one:

Hint: You can fill first two places in anyway you want , that is 10^2 ways , and then the last one in 9 or 8 ways .Add both cases to get the required probability and divide them by / 10^3.

BJ