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if a 3 digit number is randomly chosen what is the probability that either the number or some permutation of the number (which is a  3 digit number ) is divisible by 4 and 5
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it is .05
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it is 29/180
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If a number is divisible by 4 and 5, then it is divisible by 20  (since 4 and 5 are relatively prime)….

A number is divisible by 20 if it ends in 20,40,60,80 or 00....

There exist two cases...

Numbers ending in 20,40,60,80, and numbers ending in 00…

 

      I.  Number of three-digit numbers divisible by 20 ending in 20,40,60,80 =4×9 =36...

Possible permutations of each of these three-digit number, including the number itself

=4 (e.g. 520→502,205,250) ...

Therefore, numbers in this case = 36×4 =144 ...

However, when both non-zero digits are even, the permutations are repeated.

(e.g. 820→802,208,280 and 280→802,208,820)

Such numbers =24×4​=8 …

 

Subtract these cases. ⇒ total numbers in this case =144−8 =136 ..

 

  1. For those numbers ending in 00, no other permutations are possible.

Hence, numbers in this case =9 ..

Total cases

=136+9=145 ...

 

Total 3-digit numbers

=9×10×10=900 ....

Probability

=( 145​/900 ) = ( 29 / 180 ) .…

 

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1 comment

avoid  copy paste while answering,  

case 1    with two “00” total 9 such numbers are possible (100 ,200..900)

case 2   with one  0  at least one digit should be even  so 3 sub cases

           a)both digit are even and distinct (4x3)x4 =48 such numbers (204,240,420,402 4 permutation)

           b) both digits are even and same (4x1)x2=8  (e.g 220 and 202 only 2 permutation)

           c) one digit even other odd  (5x4)x4 =80  (104,140,410,401 4 permutation)

total 9+48+8+80=145 such numbers

prob =145/900 =29/180
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