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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!