A group of 30 dice are thrown. What is the probability that 5 of each of the values 1, 2, 3, 4, 5, 6 appear

the probability that 5 of each of the values 1, 2, 3, 4, 5, 6 appear is 0.0004

Step-by-step explanation:

Given that;

group of dice thrown (n) = 30

number of samples n(S) = 6³⁰

the probability that 5 of each of the values 1, 2, 3, 4, 5, 6 appear = ?

let A rep the event that 5 out of each of the values 1, 2, 3, 4, 5, 6 appear.

number of ways that 30 dice can be thrown is 30!

now, In the thirty throws from 1 to 30, there are;

FIVE 1's, FIVE 2's, FIVE 3's, FIVE 4's, FIVE 5's, FIVE 6's  characters to assign the dice.

which means n(A) = 30! / 5!5!5!5!5!5!

Now the probability that 5 of each of the values 1, 2, 3, 4, 5, 6 appear will be;

P(A) = 30!/5!5!5!5!5!5! / 6³⁰

P(A) = 30! / (5!)⁶ (6³⁰)

P(A) = 0.0004

Therefore the probability that 5 of each of the values 1, 2, 3, 4, 5, 6 appear is 0.0004

b

2

−

b

b

3

+

7

b

2

+

14

b

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6

%

−

3

b

3

+

7

b

2

+

14

b

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6

%

step-by-step explanation:

14.3

step-by-step explanation:

add all the numbers together.

9 + 12 + 15 + 16 + 17 + 17 = 86

now, divide 86 by how many numbers we have: 6.

86 / 6 = 14.3