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Hi! I am trying to figure out why the answer to the following question is 1/900,000:

 

"What is the probability of guessing my seven-digit number in ten attempts (assume a positive number that's regenerated after each guess, and no leading zero)?"

 

How do I go about doing this problem?

 Nov 27, 2020
edited by WillyGolden  Nov 27, 2020
 #1
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9 999 999 - 1 000 000 +1 = 9 000 000  seven digit numbers

ten guesses

   10 / 9 000 000 = 1 / 900 000

 Nov 27, 2020
 #2
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So I guess my follow up would be why do we start with "9 999 999 - 1 000 000 +1 = 9 000 000" to get the denominator? My guess is that you subtract the highest possible 7 digit number from the lowest, and then add one. But, I'm not sure why you add "1." Thank you for your help so far!

WillyGolden  Nov 27, 2020
 #4
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....because you subtract one of the possible numbers

     example :  two digit numbers

                          10  through 99   is   90 numbers

                                 99-10  +1 = 90 numbers      you have to add 1 because you subtracted the '10'  from the possibles....

Guest Nov 27, 2020
 #3
avatar+31703 
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The probability of getting the first digit right in one try is 1/9 (because there are only 9 possibilities, since we can't have a leading zero).

 

The probability of getting the second digit right, having got the first one right is (1/9)*(1/10) 

 

The probability of getting the third digit right, having got the first two right is (1/9)*(1/10)2 

...

...

 The probability of getting the seventh digit right, having got the first six right is (1/9)*(1/10)6 

 

If you have ten independent attempts then the probability is 10*(1/9)*(1/10)6  or (1/9)*(1/10)5  or 1/900,000

 Nov 27, 2020

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