Four tickets marked $00,01,10$ and $11$ respectively are placed in a bag. A ticket is drawn at random five times, being replaced each time. The probability that the sum of the numbers on the tickets is $15$, is
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$\displaystyle \frac { 3 }{ 1024 } $
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$\displaystyle \frac { 5 }{ 1024 } $
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$\displaystyle \frac { 7 }{ 1024 } $
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None of these
B
Correct answer
Explanation
Each ticket is a binary number: 0, 1, 2, 3. We draw 5 times with replacement. Total outcomes = 4^5 = 1024. We want the sum of 5 draws to be 15. This is equivalent to finding the number of integer solutions to x1+x2+x3+x4+x5 = 15 where 0 <= xi <= 3. Using generating functions, the coefficient of x^15 in (1+x+x^2+x^3)^5 is 5.