Multiple choice

A die is thrown $ 2n $ times. The probability that the number greater than 4 appears at least once in $ 2n $ throws is

  1. $ \displaystyle \left ( \frac{1}{3} \right )^{2n} $
  2. $ 1-\displaystyle \left ( \frac{1}{3} \right )^{2n} $
  3. $ \displaystyle \frac{3^{2n}-\:2^{2n}}{3^{2n}} $

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The probability of getting a number greater than 4 (i.e., 5 or 6) in one throw is 2/6 = 1/3. The probability of not getting a number greater than 4 is 1 - 1/3 = 2/3. In 2n throws, the probability of never getting a number greater than 4 is (2/3)^(2n). Thus, the probability of getting at least one is 1 - (2/3)^(2n) = 1 - (2^(2n) / 3^(2n)) = (3^(2n) - 2^(2n)) / 3^(2n).

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