Hi ,
I have a probability problem and would want to know the solution .
Problem:-
I want to create all 5 digit numbers with the below condition.
The 1st digit should be a even number. The second digit should be a odd number.
The 3rd digit should be a non even prime number. Fourth and fifth should be random digits which do not occur previously.
As per the solution the answer given is 4(even numbers)* 5( odd numbers) * 3 * 7 * (non used digits) * 6 (non used digits) which is 2520.
But here the cases have not been considered when 2nd and 3rd digit can be the same ie the below cases (3,3) , (5,5) and (7,7). In these cases the 4th and 5th numbers should be 8(non used digits)*7(non used digits) . Since there are 3 such combinations we multiply it by 3 ie 56*3=168.
Could anyone let me know if (2520+168) is correct or the answer 2520 is the right one
Digit Problem on Probability
This topic has expert replies
I solved it myself
Here is my solution:
Digit 1 can be {2,4,6,8}
Digit 2 can be {1,3,5,6,7}
Digit 3 can be {3,5,7}
Digit 4&5 are dependent on previous digits.
Case 1: Digit 2 and 3 are same (and belong to the subset {3,5,7})
Digit 1 can be {2,4,6,8} --> 4 options
Digit 2 can be {3} --> 1 option
Digit 3 can be {3} --> 1 option
Digit 4 can have 10-2 = 8 options
Digit 5 can have 10-3 = 7 options
---- total = 4*1*1*8*7 = 4*56 options
You can repeat case 1 three times for each 3,5 and 7 (i.e. digit 2 can be 3, 5 or 7 etc..).
Hence total options from case 1 = 3*4*56 = 12*56
Case 2: Digit 2 and 3 are different (and belong to the subset {3,5,7})
Digit 1 can be {2,4,6,8} --> 4 options
Digit 2 can be {3} --> 1 option
Digit 3 can be {5,7} --> 2 option
Digit 4 can have 10-3 = 7 options
Digit 5 can have 10-4 = 6 options
---- total = 4*1*2*7*6 = 6*56 options
You can repeat case 2 three times for each 3,5 and 7. (i.e. digit 2 can be 3, 5 or 7 etc..)
Hence total options from case 1 = 3*6*56 = 18*56
Case 3: Digit 2 belongs to the subset {1,9} and Digit 3 belongs to the subset {3,5,7}
Digit 1 can be {2,4,6,8} --> 4 options
Digit 2 can be {1,9} --> 2 option
Digit 3 can be {3,5,7} --> 3 option
Digit 4 can have 10-3 = 7 options
Digit 5 can have 10-4 = 6 options
---- total = 4*2*3*7*6 = 18*56 options
There is no repetition for this case. Total = 18*56
Grand total of possibilities are the sum of all the above totals:
GT = 12*56 + 18*56 + 18*56 = 48*56 = 2688
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Hope that helps
Here is my solution:
Digit 1 can be {2,4,6,8}
Digit 2 can be {1,3,5,6,7}
Digit 3 can be {3,5,7}
Digit 4&5 are dependent on previous digits.
Case 1: Digit 2 and 3 are same (and belong to the subset {3,5,7})
Digit 1 can be {2,4,6,8} --> 4 options
Digit 2 can be {3} --> 1 option
Digit 3 can be {3} --> 1 option
Digit 4 can have 10-2 = 8 options
Digit 5 can have 10-3 = 7 options
---- total = 4*1*1*8*7 = 4*56 options
You can repeat case 1 three times for each 3,5 and 7 (i.e. digit 2 can be 3, 5 or 7 etc..).
Hence total options from case 1 = 3*4*56 = 12*56
Case 2: Digit 2 and 3 are different (and belong to the subset {3,5,7})
Digit 1 can be {2,4,6,8} --> 4 options
Digit 2 can be {3} --> 1 option
Digit 3 can be {5,7} --> 2 option
Digit 4 can have 10-3 = 7 options
Digit 5 can have 10-4 = 6 options
---- total = 4*1*2*7*6 = 6*56 options
You can repeat case 2 three times for each 3,5 and 7. (i.e. digit 2 can be 3, 5 or 7 etc..)
Hence total options from case 1 = 3*6*56 = 18*56
Case 3: Digit 2 belongs to the subset {1,9} and Digit 3 belongs to the subset {3,5,7}
Digit 1 can be {2,4,6,8} --> 4 options
Digit 2 can be {1,9} --> 2 option
Digit 3 can be {3,5,7} --> 3 option
Digit 4 can have 10-3 = 7 options
Digit 5 can have 10-4 = 6 options
---- total = 4*2*3*7*6 = 18*56 options
There is no repetition for this case. Total = 18*56
Grand total of possibilities are the sum of all the above totals:
GT = 12*56 + 18*56 + 18*56 = 48*56 = 2688
---
Hope that helps