# Discrete Math - quick probability questions.

## Main Question or Discussion Point

For the life of me I am having a hard time understanding how to do problems of this nature. As I understand it, were using the multiplication rule here with a twist.

a. How many integers from 1 through 100,000 contain the
digit 6 exactly once?

5 * 9 * 9 * 9 * 9 = 38805 is what I have. Because the digit 6 can appear in 5 different locations.

b. How many integers from 1 through 100,000 contain the
digit 6 at least once?

c. If an integer is chosen at random from 1 through
100,000, what is the probability that it contains two or
more occurrences of the digit 6?

I am clueless here. Add up all the possibilities of containing two 6's...three 6's... four 6's... five 6's and then divide by the total possibilities.That's my line of thinking... I am just not sure how to find the amount of possibilities for each partition.

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pwsnafu
b. How many integers from 1 through 100,000 contain the
digit 6 at least once?
How many integers are there that do not contain any digit 6?

• KingsFoil
b. How many integers from 1 through 100,000 contain the
digit 6 at least once?
The first thing I always try in probability problems is to invert it. P(X) = 1 - P(not X)
It's easy to give it a shot, and often the resulting problem is much simpler.

In this case not X is
b') How many integers from 1 through 100,000 do not contain digit 6?
I bet that you can do that.

c. If an integer is chosen at random from 1 through
100,000, what is the probability that it contains two or
more occurrences of the digit 6?

I am clueless here. Add up all the possibilities of containing two 6's...three 6's... four 6's... five 6's and then divide by the total possibilities.That's my line of thinking... I am just not sure how to find the amount of possibilities for each partition.
Try my special magic rule again.

• KingsFoil and fresh_42