Probability - Inclusion - Exlcusion

Probability Level 3

A number is selected at random from S = { 1000 , 1001 , . . . . , 9999 } S=\{1000, 1001, ...., 9999\} .
If the probability

Pr( the selected number has at least one digit that is 0, and at least one digit is 1 and at least one digit is 2) = a b = \cfrac{a}{b}

where a a and b b are co-prime integers.

Find a + b a+b .

Example:

  • 1082 1082 and 2102 2102 are some of the desired numbers.
  • but 1234 1234 and 2069 2069 are not the desired numbers.


The answer is 61.

Pop Wong by Pop Wong , 47, Hong Kong

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1 solution

Mark Hennings
Aug 16, 2020

The digits of the number could be 0 , 0 , 1 , 2 0,0,1,2 or 0 , 1 , 1 , 2 0,1,1,2 or 0 , 1 , 2 , 2 0,1,2,2 or 0 , 1 , 2 , x 0,1,2,x where x ∈ { 3 , 4 , 5 , 6 , 7 , 8 , 9 } x \in \{3,4,5,6,7,8,9\} .

  • there are 3 × 2 = 6 3\times2 = 6 numbers of the first type,
  • there are 3 × 3 = 9 3 \times3 = 9 numbers of the second type,
  • there are 3 × 3 = 9 3 \times 3 = 9 numbers of the third type,
  • there are 7 × 3 × 3 ! = 126 7 \times 3 \times 3! = 126 numbers of the fourth type,

where the above counts ensure that no number begins with the digit 0 0 . Thus the desired probability is 6 + 9 + 9 + 126 9000 = 1 60 \frac{6 + 9 + 9 + 126}{9000} \; = \; \frac{1}{60} making the answer 1 + 60 = 61 1+60 = \boxed{61} .

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