Lewis is rolling a 10-sided die 250 times. Each side of the die is labeled with a number 1 – 10. Based on the theoretical probability, how many times would Lewis be expected to roll a multiple of 3? 25 30 75 83

Respuesta :

okay so multiples of 3 on a 10 sided die is 3,6,and 9 so 3/10 times 
if your rolling it 250 times you cross multiply with x so the equation would be 3/10= x/250
do 3x250 which is 750 
750//10 is 75
x=75 so its your answer:)

Using the binomial distribution, Lewis would be expected to roll a multiple of 3 83 times.

For each roll, there are only two possible outcomes, either a multiple of 3 is rolled, or it is not. The probability of a multiple of 3 being rolled on a trial is independent of any other trial, hence the binomial distribution is used.

What is the binomial probability distribution?

It is the probability of exactly x successes on n repeated trials, with p probability of a success on each trial.

The expected value of the binomial distribution is:

E(X) = np

In this problem:

  • The die is rolled 250 times, hence n = 250.
  • Of the 10 numbers, 3 of them are multiples of 3(3, 6 and 9), hence p = 3/10 = 0.3.

Then:

E(X) = np = 250(0.3) = 83.3.

Rounding, Lewis would be expected to roll a multiple of 3 83 times.

You can learn more about the binomial distribution at https://brainly.com/question/14424710