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A dodecahedral die (one with 12 sides numbered from 1 to 12) is tossed once. Find the probability that the number on the upward face is divisible by 3. (Enter the probability as a fraction.)

Sagot :

allaliar

Answer:

the probability is [tex]\frac{1}{3}[/tex]

Step-by-step explanation:

there are 4 numbers on a 12 sided die that can be divided by 3: 3, 6, 9, and 12.

so the probability would be [tex]\frac{4}{12}[/tex], or [tex]\frac{1}{3}[/tex]

fichoh

Using the concept of probability. The chances of having a value which is divisible by 3 is 1/3

Recall :

  • Probability = required outcome / Total possible outcomes

Total possible outcomes = Sample space = 12

Required outcome = numbers divisible by 3 :

  • (3, 6, 9, 2) = 4 outcomes

P(number divisible by 3) = 4 / 12 = 1/3

Therefore, the probability of having a number which is divisible by 3 is 1/3

Learn more : https://brainly.com/question/18405415

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