An ordinary die is rolled 30 times, how many times would you expect the result to be a six?

Question

I would like to know the probability in a fraction of these two questions...

1. An ordinary die is rolled 30 times, how many times would you expect the result to be a six?

2.An ordinary die is rolled 30 times, how many times would you expect the result to be a number greater than 2?

Answer 1

The expected value of these situations is simply the number of rolls times the probability of each specific outcome.

Problem 1) The probability of rolling a 6 is 1/6 since there is one 6 out of the six sides. The expected number of 6's would be 30 * (1/6) = 30/6 = 5 times.

Problem 2) The probability of rolling a number greater then 2 is 2/3 since there are four numbers greater than 2 out of the six sides. The expected number of values greater than 2 would be 30*(2/3) = 60/3 = 20 times.

Answer 2

Part 1)

Each roll of a die is an independent event. So, you would like to add the probability of a 6 on each throw:

(1/6) + (1/6) + (1/6) + (1/6) + (1/6) + (1/6) + (1/6) + ... 30 times

= (1/6) *30

= 30/6

=5.

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Part 2)

Probability that the number will be greater than 2 = 4/6.

Repeat the analysis in Part 1.

(4/6) * 30

= 20

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