Question

In: Statistics and Probability

You roll 3 standard fair dice. Let A be the event that the sum is odd....

You roll 3 standard fair dice. Let A be the event that the sum is odd. Let B be the event that the sum is more than 14. Let C be the event that the sum is more than 9.
All probabilities to the thousandths place.

Find the probability of the basic events: A, B, C
Find the probability of the intersection events: A∩B, A∩C, B∩C, A∩B∩C
Find the probability of the union events:AUB,AUC,BUC,AUBUC
Find the probability of the conditional events: (A|B), (B|A), (A|C), (C|A), (B|C), (C|B)
Two of the above probabilities are equal to one another. (There are 2 possible correct

answers) Identify them and in a sentence explain why they are the same.
One of the above probabilities is equal to 1. Identify it and in a sentence explain why it has

to be equal to 1.

Expert Solution

Probability = Favorable Outcomes / Total Outcomes

Since 3 die are thrown, Total outcomes = 6³ = 216

Since the sum can be either odd or even, there will be 108 outcomes where the sum is even and 108 outcomes, where the sum is odd.

Therefore P(A) = 108 / 216 = 0.500

____________

Sum of More than 14 = Sum of 15, 16 17 or 18

Outcomes where a Sum of 15 occurs = (3 6 6), (6 3 6), (6 6 3), (4 5 6), (4 6 5), (5 4 6), (5 6 4), (6 4 5), (6 5 4), (5 5 5) = 10

Outcomes where a Sum of 16 occurs = (4 6 6), (6 4 6), (6 6 4), (5 5 6), (5 6 5), (6 5 5) = 6

Outcomes where a Sum of 17 occurs = (5 6 6), (6 5 6), (6 6 5) = 3

Outcomes where a Sum of 18 occurs = (6 6 6) = 1

Therefore Favorable Outcomes = 10 + 6 + 3 + 1 = 20

Therefore P(B) = 20 / 216 = 0.093

_________________

Sum greater than 9 = Sum of 10, 11, 12, 13, 14 , 15, 16, 17, 18

We know to get a sum of 15, 16, 17 or 18, there are 20 outcomes

Outcomes where a Sum of 10 occurs = (1 3 6), (1 6 3), (3 1 6), (3 6 1), (6 1 3), (6 3 1), (1 4 5), (1 5 4) (4 1 5), (4 5 1), (5 1 4), (5 4 1), (2 2 6), (2 6 2), (6 2 2), (2 3 5), (2 5 3), (3 2 5), (3 5 2), (5 2 3), (5 3 2), (2 4 4), (4 2 4), (4 4 2), (3 3 4), (3 4 3) (4 3 3) = 27

Outcomes where a Sum of 11 occurs = (1 4 6), (1 6 4), (4 1 6), (4 6 1), (6 1 4) (6 4 1), (1 5 5), (5 1 5), (5 5 1), (2 3 6), (2 6 3), (3 2 6), (3 6 2), (6 2 3) (6 3 2), (2 4 5), (2 5 4), (4 2 5), (4 5 2), (5 2 4) (5 4 2) (3 3 5), (3 5 3), (5 3 3) (3 4 4), (4 3 4), (4 4 3) = 27

Outcomes where a Sum of 12 occurs = (1 5 6), (1 6 5), (5 1 6), (5 6 1), (6 1 5) (6 5 1), (2 4 6), (2 6 4), (4 2 6), (4 6 2), (6 2 4) (6 4 2), (2 5 5), (5 2 5), (5 5 2), (3 3 6), (3 6 3) (6 3 3) (3 4 5), (3 5 4), (4 3 5) (4 5 3), (5 3 4), (5 4 3) (4 4 4) = 25

Outcomes where a Sum of 13 occurs = (1 6 6), (6 1 6), (6 6 1), (2 5 6), (2 6 5) (5 2 6), (5 6 2), (6 2 5), (6 5 2), (3 4 6), (3 6 4) (4 3 6), (4 6 3), (6 3 4), (6 4 3), (3 5 5), (5 3 5) (5 5 3) (4 4 5), (4 5 4), (5 4 4) = 21

Outcomes where a Sum of 14 occurs = (2 6 6), (6 2 6), (6 6 2), (3 5 6), (3 6 5) (5 3 6) (5 6 3), (6 5 3), (6 3 5) (4 4 6), (4 6 4), (6 4 5) (4 5 5) (5 4 5) (5 5 4) = 15

Favorable Outcomes for sum greater than 9 = 27 + 27 + 25 + 21 + 15 + (20) = 135

Therefore P(C) = 135 / 216 = 0.625

___________________________

P(A B)

Favorable Outcomes = Sum is ODD and Sum greater than 14 = Sum of 15 and 17 = 10 + 3 = 13

Therefore P(A B) = 13 / 216 = 0.060

___

P(A C)

Favorable Outcomes = Sum is ODD and Sum greater than 9 = Sum of 11, 13, 15 and 17 = 27 + 21 + 10 + 3 = 61

Therefore P(A C) = 61 / 216 = 0.282

___

P(B C)

Favorable Outcomes = Sum is greater than 14 and Sum greater than 9 = Sum is greater than 14 = Sum of 15, 16, 17 and 18 = P(B)

Therefore P(B C) = P(B) = 0.093

___

P(A B C)

Favorable Outcomes = Sum is ODD and Sum greater than 9 and Sum is Greater than 14 = Sum of 15 and 17 = P(A B)

Therefore P(A B C) = P(A B) = 0.060

___________________________

P(A U B) = P(A) + P(B) - P(A B) = 0.5 + 0.093 - 0.06 = 0.533

P(A U C) = P(A) + P(C) - P(A C) = 0.5 + 0.625- 0.282 = 0.843

P(B U C) = P(B) + P(C) - P(B C) = 0.093 + 0.625 - 0.093 = 0.625

P(A U B U C) = P(A) + P(B) + P(C) - P(A B) - P(A C) - P(B C) + P(A B C) = 0.5 + 0.093 + 0.625 - 0.06 - 0.282 - 0.093 + 0.06 = 0.843

_________________________________

By Bayes Theorem P(A / B) = P(A given B) = P(A / B) = P(A B) / P(B)

Please note P(A B) = P(B A)

P(A / B) = P(A B) / P(B) = 0.06 / 0.093 = 0.645

P(B / A) = P(A B) / P(A) = 0.06 / 0.5 = 0.120

P(A / C) = P(A C) / P(C) = 0.282/ 0.625 = 0.451

P(C / A) = P(A C) / P(A) = 0.282 / 0.5 = 0.564

P(B / C) = P(B C) / P(C) = 0.093 / 0.625 = 0.149

P(C / B) = P(B C) / P(B) = 0.093 / 0.093 = 1.000

__________________________________

The 2 Probabilities which are the Same:

(1) P(B C) = P(B) They are equal as the event B is also included in C, and therefor the common event is B.

(2) P(A B C) = P(A B). They are equal as A B (Sum is Odd and Sum greater than 14) is a part of A C sum is odd and sum greater than 9. Therefore (A B) (A C) = A B

___________________

The probability which is equal to 1 is P(C / B) = P(B C) / P(B). This is because P(B C) = P(B). hey are equal as the event B is also included in C, and therefor the common event is B.

____________________

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orchestra answered 1 year ago

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You roll 3 standard fair dice. Let A be the event that the sum is odd....

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