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General
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Question

Four tickets numbered 00, 01, 10, 11 are placed in a bag A ticket is drawn at random and replaced Again a ticket is drawn at random The probability that the sum of the numbers on the tickets drawn is 21

  1. fraction numerator 1 over denominator 4 end fraction    
  2. fraction numerator 1 over denominator 8 end fraction    
  3. fraction numerator 1 over denominator 6 end fraction    
  4. fraction numerator 1 over denominator 12 end fraction    

The correct answer is: fraction numerator 1 over denominator 8 end fraction


    Favorable cases equals{ left parenthesis 10 comma blank 11 right parenthesis left parenthesis 11 comma blank 10 right parenthesis } n left parenthesis E right parenthesis equals 2 n left parenthesis S right parenthesis equals 4 to the power of 2 end exponent equals 16
    P left parenthesis E right parenthesis equals fraction numerator n left parenthesis E right parenthesis over denominator n left parenthesis S right parenthesis end fraction equals fraction numerator 2 over denominator 16 end fraction equals fraction numerator 1 over denominator 8 end fraction

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