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## Comment on

4-Digit Number## don't you think the 2nd, 3rd

## No, it's still 1/10 (although

No, it's still 1/10 (although it certainly seems like it might be 1/9).

However, it doesn't matter that the 1st value can't be 0. There are 10 possible values for the 2nd digit, and only 1 of them matches the 1st digit. So, the probability that the 2nd number matches the first is 1/10

## For probability approach: Why

## We're selecting an integer

We're selecting an integer from 1000 to 9999, so it's impossible for the 1st digit (the thousands digit) to be zero.

## Hi Brent, since we're

## Can you tell me what 9/10

Can you tell me what 9/10 refers to?

## Sure Brent.

Since we're selecting an integer from 1000 to 9999, so first digit (0 ~ 9) but start from 1 (so 0 is out). Therefore we are selecting the rest of 9 digits out of total 10 digits (0 ~ 9) > 9/10? Could you help clarify? Thanks Brent

## Sorry but I'm not sure what

Sorry but I'm not sure what you mean by "selecting the rest of 9 digits out of total 10 digits."

We aren't selecting 9 digits here. We are selecting 4 digits: The thousands digit, the hundreds digit, the tens digit, and the ones digit.

If we are counting the number of 4-digit numbers with all 4 digits identical, then:

There are 9 ways to select the thousands digit (1 to 9)

Important: Once we have selected the thousands digit, the remaining three digits must match the thousands digit.

So we have:

There is 1 way to select the hundreds digit (it must match the thousands digit)

There is 1 way to select the tens digit (it must match the thousands digit)

There is 1 way to select the ones digit (it must match the thousands digit)

So, the total number of 4-digit numbers with all digits the same = 9 x 1 x 1 x 1 = 9

There are 9000 integers from 1000 to 9999

So the probability of selecting a 4-digit number where all of the digits are the same = 9/9000

Does that help?

## Get it thanks Brent

## Can we imply that the

## Yes, that would be an

Yes, that would be an equivalent model.

Great idea!

## Why isn't the probability of

## The first digit can be ANY

The first digit can be ANY digit among 1, 2, 3, 4, 5, 6, 7, 8, and 9

It doesn't matter which digit it is. What matters is that the remaining digits match the first digit.

Does that help?

## Why is the probability of the

## If we want all 4 digits to be

If we want all 4 digits to be the same, then it doesn't matter what the first digit is.

For example, if the first digit is a 7 then, in order for all 4 digits to be the same, we need the other three digits to be 7.

Similarly, if the first digit is a 9 then, in order for all 4 digits to be the same, we need the other three digits to be 9.

If the first digit is a 2 then, in order for all 4 digits to be the same, we need the other three digits to be 2.

So, the likelihood of all 4 digits being the same does not depend on the value of the first digit.

In fact, the first digit can be ANY digit.

If we randomly choose a digit to be the first digit, what is the probability that the selected digit is ANY digit?

Since our first selection is guaranteed to be an integer, the probability is 1.

Does that help?

Cheers,

Brent

## I have this question in mind

## Each digit is RANDOMLY

Each digit is RANDOMLY selected.

So, for example, if we select 7 for the first digit, this has no affect on the likelihood of randomly selecting a 7 for the second digit.

So, those two events are independent.

We can apply the same logic to show that all 4 events (selecting the 4 digits) are independent.

Does that help?

Cheers,

Brent

## This definitely helps. Thanks

## Hi!

My approach:

Total outcomes = 9000

favourable outcomes: 9 (1111,2222,3333 and so on up to 9999)

hence probability = 9/9000 = 1/1000

## That's a perfect approach (it

That's a perfect approach (it's also the same as the first solution in the video :-)

## Is the following a valid

## Great approach. Nice work!

Great approach. Nice work!