Probability of Sums from Rolling Two Dice: At Most 3, Exactly 6, At Least 9

Math Problem Statement

A standard die has 6 sides numbered 1 through 6. What is the best and most specific way of computing the probabilities of outcomes from rolling a pair of standard dice?

one-step fraction or division:

number specified

total number

list or fraction or division of pairs:

number specified pairs

total pairs

use tree or multiplication rule:

P(A ∩ B) = P(A | B) · P(B)

box, Venn diagram, complement rule:

1-P(A)

, or addition rule:

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

conditional probability formula:

P(A|B) =

P(A∩ B)

P(B)

permutation(s):

(n−k)!

combination(s):

(n−k)k!

other factorial(s):

exponent(s):

What is the probability that the sum of the numbers showing is: (a) at most 3 ? Write answer as a reduced fraction.

(b) exactly 6 ? Write answer as a reduced fraction.

Solution

To analyze the probability of outcomes from rolling a pair of standard dice, we need to identify the total number of possible pairs and then count the number of specific pairs that meet the criteria for each condition.

Step-by-Step Analysis

When two six-sided dice are rolled, each die has 6 outcomes. Thus, the total number of outcomes is: $6 \times 6 = 36$ This represents all possible ordered pairs of the numbers 1 through 6, where the first number is the result from the first die and the second number is the result from the second die.

Probability Calculations

For each condition, we will find the number of successful pairs and divide it by 36, the total number of pairs.

(a) Probability that the sum is at most 3

We need to count the pairs $(a, b)$ such that $a + b \leq 3$ .

Possible pairs: $(1,1)$ , $(1,2)$ , $(2,1)$

Number of successful pairs = 3.

Thus, the probability is: $\frac{3}{36} = \frac{1}{12}$

(b) Probability that the sum is exactly 6

We need to count the pairs $(a, b)$ such that $a + b = 6$ .

Possible pairs: $(1,5)$ , $(2,4)$ , $(3,3)$ , $(4,2)$ , $(5,1)$

Number of successful pairs = 5.

Thus, the probability is: $\frac{5}{36}$

(c) Probability that the sum is at least 9

We need to count the pairs $(a, b)$ such that $a + b \geq 9$ .

Possible pairs: $(3,6)$ , $(4,5)$ , $(4,6)$ , $(5,4)$ , $(5,5)$ , $(5,6)$ , $(6,3)$ , $(6,4)$ , $(6,5)$ , $(6,6)$

Number of successful pairs = 10.

Thus, the probability is: $\frac{10}{36} = \frac{5}{18}$

Final Answers

(a) Probability that the sum is at most 3: $\frac{1}{12}$
(b) Probability that the sum is exactly 6: $\frac{5}{36}$
(c) Probability that the sum is at least 9: $\frac{5}{18}$

Would you like further details on these calculations, or have any questions?

Tip

Always start by identifying the total number of possible outcomes and then systematically count the number of successful outcomes for your specific condition. This approach helps in structuring probability problems clearly.

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Math Problem Analysis

Mathematical Concepts

Probability
Combinatorics

Formulas

P(A ∩ B) = P(A | B) · P(B)
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
Conditional probability formula: P(A | B) = P(A ∩ B) / P(B)

Theorems

Multiplication rule
Addition rule
Complement rule

Suitable Grade Level

Grades 6-8

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