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A four-digit natural number "abcd" is "balanced" if a + d = b + c. What is the probability that it is a "balanced" if a four-digit natural number is randomly selected?

edited by aboslutelydestroying  Apr 7, 2024
edited by aboslutelydestroying  Apr 7, 2024

Best Answer 

 #1
avatar+9664 
+1

I could not figure out a nice way to calculate the number of "balanced" numbers, but I wrote a Python program to count them.

The output of the program tells me that there are 615 "balanced" numbers. Hence, the required probability is \(\dfrac{615}{9000} = \dfrac{41}{600}\).

 

Would love to see a combinatorial approach to this problem.

 Apr 8, 2024
 #1
avatar+9664 
+1
Best Answer

I could not figure out a nice way to calculate the number of "balanced" numbers, but I wrote a Python program to count them.

The output of the program tells me that there are 615 "balanced" numbers. Hence, the required probability is \(\dfrac{615}{9000} = \dfrac{41}{600}\).

 

Would love to see a combinatorial approach to this problem.

MaxWong Apr 8, 2024
 #2
avatar+214 
+1

Thankssmiley


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