Multiple choice

A fair die is tossed until a number greater than $4$ appear. The probability that an even number of tosses shall be required is:

  1. $ \displaystyle \frac{1}{2} $
  2. $ \displaystyle \frac{3}{5} $
  3. $ \displaystyle \frac{2}{5} $
  4. $ \displaystyle \frac{2}{3} $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Probability of success (number > 4) is 2/6 = 1/3. Probability of failure is 2/3. Even tosses means success on 2nd, 4th, 6th... toss. P = (2/3)(1/3) + (2/3)^3(1/3) + ... = (2/9) / (1 - 4/9) = (2/9) / (5/9) = 2/5.

AI explanation

The probability of rolling a number greater than 4 is 2/6, which simplifies to 1/3, meaning the probability of rolling a number less than or equal to 4 is 2/3. To find the probability that a number greater than 4 appears on an even-numbered toss, we sum the infinite geometric series for the 2nd, 4th, 6th, and subsequent even tosses. This series is (2/3)(1/3) + (2/3)^3(1/3) + (2/3)^5*(1/3), which evaluates to a sum of 2/5.

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