A fair ten-sided dice has faces numbered 1-10. the dice is rolled five times. a) on each roll, how many ways are there to roll a multiple of 3? b) how many different ways are there to roll a multiple of 3 on all five rolls? c) what is the probability of rolling a multiple of 3 on all five rolls?



Answer :

Using the Fundamental Counting Theorem, the measures are given as follows:

a) 3 ways to roll a multiple of 3 on each trial.

b) 243 ways to roll a multiple of 3 on all five trials.

c) 0.00243 probability of rolling a multiple of 3 on all five rolls.

What is the Fundamental Counting Theorem?

It is a theorem that states that if there are n independent trials, each with [tex]n_1, n_2, \cdots, n_n[/tex] possible results, the total number of outcomes is calculated by the multiplication of the factorials as presented as follows:

[tex]N = n_1 \times n_2 \times \cdots \times n_n[/tex]

The multiples of 3 between 1 and 10 are given as follows:

3, 6 and 9.

Hence three multiples, which is the number of ways to roll a multiple of 3 on each trial.

Hence, the number of ways to roll a multiple of 3 on all five trials is:

N = 3 x 3 x 3 x 3 x 3 = 3^5 = 243.

The total number of outcomes of the five trials is:

T = 10^5.

Then the probability of rolling a multiple of 3 on all five rolls is:

243/(10^5) = 0.00243.

More can be learned about the Fundamental Counting Theorem at https://brainly.com/question/15878751

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