Multiple choice

Three faces of an ordinary dice are yellow, two faces are red and one face is blue. The dice is tossed $3$ times. The probability that yellow, red and blue faces appear in the first, second and third tosses respectively is

  1. $\displaystyle \frac {1}{36}$
  2. $\displaystyle \frac {1}{6}$
  3. $\displaystyle \frac {1}{30}$

  4. none of these
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A Correct answer
Explanation

P(Yellow) = 3/6 = 1/2. P(Red) = 2/6 = 1/3. P(Blue) = 1/6. The probability of the specific sequence (Yellow, Red, Blue) is (1/2) * (1/3) * (1/6) = 1/36.

AI explanation

The probability of independent events occurring in sequence is the product of their individual probabilities. The probability of rolling yellow is 3/6, red is 2/6, and blue is 1/6. Multiplying these gives (3/6) * (2/6) * (1/6) = 6/216, which simplifies to 1/36.

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