Multiple choice

A coin whose faces are marked 3 & 5 is tossed 4 times. The probability that the sum of the numbers thrown is 12 is

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

Possible sums with 4 tosses (each 3 or 5): 3+3+3+3=12. This is the only combination that sums to 12. The number of ways to get 3,3,3,3 is 1. Total outcomes = 2^4 = 16. Probability = 1/16.

AI explanation

To obtain a sum of 12 from four tosses of a coin marked 3 and 5, the coin must land on 3 every time, since 4 times 3 equals 12. The probability of getting a 3 on a single toss is 1 out of 2. By the multiplication theorem for independent events, the probability of getting four 3s in a row is (1/2) raised to the power of 4, which equals 1 divided by 16.

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