How do you solve probability problems
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How do you solve probability problems

[From: ] [author: ] [Date: 12-04-03] [Hit: ]
-9 digits, so there are 10^9 possibilities in total.For part (b), the easier way is to find how many do NOT have a repeated digit, and subtract that number from 10^9.10 possibilities for the first digit,......
I keep getting the wrong answer...

A ssn number is in this form xxx-xx-xxxx where x is a digit from 0 to 9:

A) how many ssn have exactly 4 nines?
B) how many with at least one repeated #?

-
9 digits, so there are 10^9 possibilities in total.

For part (b), the easier way is to find how many do NOT have a repeated digit, and subtract that number from 10^9. To get no repeated digit you have:
10 possibilities for the first digit,
then 9 possibilities for the 2nd digit,
then 8 possibilities for the 3rd digit, etc. down to 2 possibilities for the 9th.
So the answer is 10x9x8x7x6x5x4x3x2. (Then subtract that from 10^9, remember).

question a is a bit harder. First you need to work out the number of ways you can place the four nines in your number. For 4 identical items in 9 places, it is C(9.4), or 9x8x7x6 / (4x3x2x1) = 126, you should have been taught how to do this this. Then for *each* of those 126 arrangements of the nines, for the other 5 digits you have:
9 choices for the 1st non-9 x 8 choices for the 2nd .... etc
= 9x8x7x6x5
Multiply that number by 126 to get the total.
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