Multiple choice

A coin whose faces are marked 3 & 5 is tossed 4 times. The probability that the sum of the numbers thrown is greater than 15

  1. $\displaystyle\frac{11}{16}$
  2. $\displaystyle\frac{5}{16}$
  3. $\displaystyle\frac{5}{8}$
  4. $\displaystyle\frac{1}{16}$
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A Correct answer
Explanation

The total number of outcomes is 2^4 = 16. The possible sums are: 3+3+3+3=12, 3+3+3+5=14 (4 ways), 3+3+5+5=16 (6 ways), 3+5+5+5=18 (4 ways), 5+5+5+5=20 (1 way). Sums greater than 15 are 16, 18, and 20, which occur in 6+4+1 = 11 cases.

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