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$Probablity Hw 4.75-419 Question 15 of 22 (f point) I Question Attempt: 1 of 3 7 8 ✓9 10 11 ×12 13 ✓14 15 16 17 Incorrect Your answer is incorrect. Event A. Your answer is incorrect. Event B: Your answer is incorrect. An ordinary (fair) die is a cube with the numbers 1 through 6 on the sides (represented by painted spots). Imagine that such a die is rolled twice in succes and that the face values of the two rolls are added together. This sum is recorded as the outcome of a single trial of a random experiment. Compute the probability of each of the following events. Event A: The sum is greater than 6 . Event B : The sum is divisible by 4 or 6 (or both). Write your answers as fractions. (a) P(A)=1,2,3,4,5,6 (b) P(B)= ◻ Check Save For Later submit Teman of the Pharenes$

Probablity Hw $4$ . $75$ $-419$ $\newline$ Question $15$ of $22$ (f point) I Question Attempt: $1$ of $3$ $\newline$ $7$ $\newline$ $8$ $\newline$ $\checkmark 9$ $\newline$ $10$ $\newline$ $11$ $\newline$ $\times 12$ $\newline$ $13$ $\newline$ $\checkmark 14$ $\newline$ $15$ $\newline$ $16$ $\newline$ $17$ $\newline$ Incorrect $\newline$ Your answer is incorrect. $\newline$ - Event A. Your answer is incorrect. $\newline$ - Event B: Your answer is incorrect. $\newline$ An ordinary (fair) die is a cube with the numbers $1$ through $6$ on the sides (represented by painted spots). Imagine that such a die is rolled twice in succes and that the face values of the two rolls are added together. This sum is recorded as the outcome of a single trial of a random experiment. $\newline$ Compute the probability of each of the following events. $\newline$ Event $A:$ The sum is greater than $6$ . $\newline$ Event $B$ : The sum is divisible by $4$ or $6$ (or both). $\newline$ Write your answers as fractions. $\newline$ (a) $P(A)=1,2,3,4,5,6$ $\newline$ (b) $P(B)=$ $\square$ $\newline$ Check $\newline$ Save For Later $\newline$ submit $\newline$ Teman of the $\newline$ Pharenes

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Math Problems

Algebra 2

Identify independent events

Full solution

Q. Probablity Hw

4

75

-419

\newline

Question

15

22

(f point) I Question Attempt:

1

3

\newline

7

\newline

8

\newline

\checkmark 9

\newline

10

\newline

11

\newline

\times 12

\newline

13

\newline

\checkmark 14

\newline

15

\newline

16

\newline

17

\newline

Incorrect

\newline

Your answer is incorrect.

\newline

- Event A. Your answer is incorrect.

\newline

- Event B: Your answer is incorrect.

\newline

An ordinary (fair) die is a cube with the numbers

1

through

6

on the sides (represented by painted spots). Imagine that such a die is rolled twice in succes and that the face values of the two rolls are added together. This sum is recorded as the outcome of a single trial of a random experiment.

\newline

Compute the probability of each of the following events.

\newline

Event

A:

The sum is greater than

6

\newline

Event

B

: The sum is divisible by

4

6

(or both).

\newline

Write your answers as fractions.

\newline

(a)

P(A)=1,2,3,4,5,6

\newline

(b)

P(B)=

\square

\newline

Check

\newline

Save For Later

\newline

submit

\newline

Teman of the

\newline

Pharenes

Calculate Total Outcomes: Calculate the total number of possible outcomes when rolling a die twice. Since each die has $6$ faces, the total outcomes are $6 \times 6$ .
List Sums Greater Than $6$ : List all the possible sums greater than $6$ (event $A$ ). These are $7$ , $8$ , $9$ , $10$ , $11$ , and $12$ .
Count Ways for Each Sum: Count the number of ways to get each sum from step $2$ . For $7$ : $(1,6)$ , $(2,5)$ , $(3,4)$ , $(4,3)$ , $(5,2)$ , $(6,1)$ . For $8$ : $(2,6)$ , $(3,5)$ , $(1,6)$ $0$ , $(1,6)$ $1$ , $(1,6)$ $2$ . For $(1,6)$ $3$ : $(1,6)$ $4$ , $(1,6)$ $5$ , $(1,6)$ $6$ , $(1,6)$ $7$ . For $(1,6)$ $8$ : $(1,6)$ $9$ , $(2,5)$ $0$ , $(2,5)$ $1$ . For $(2,5)$ $2$ : $(2,5)$ $3$ , $(2,5)$ $4$ . For $(2,5)$ $5$ : $(2,5)$ $6$ . Add them up.
Calculate Probability of Event A: Calculate $P(A)$ by dividing the number of favorable outcomes for event A by the total number of outcomes. There are $6$ ways to get $7$ , $5$ ways to get $8$ , $4$ ways to get $9$ , $3$ ways to get $10$ , $2$ ways to get $6$ $0$ , and $6$ $1$ way to get $6$ $2$ . So, $6$ $3$ .
Simplify Probability A: Simplify $P(A)$ . $P(A) = \frac{21}{36} = \frac{7}{12}$ .
List Sums Divisible by $4$ or $6$ : List all the sums divisible by $4$ or $6$ (event $B$ ). These are $4$ , $6$ , $8$ , $12$ .
Count Ways for Each Sum in Event B: Count the number of ways to get each sum from step $6$ . For $4$ : $(1,3)$ , $(2,2)$ , $(3,1)$ . For $6$ : $(1,5)$ , $(2,4)$ , $(3,3)$ , $(4,2)$ , $(5,1)$ . For $(1,3)$ $0$ : $(1,3)$ $1$ , $(1,3)$ $2$ , $(1,3)$ $3$ , $(1,3)$ $4$ , $(1,3)$ $5$ . For $(1,3)$ $6$ : $(1,3)$ $7$ . Add them up.
Calculate Probability of Event B: Calculate $P(B)$ by dividing the number of favorable outcomes for event B by the total number of outcomes. There are $3$ ways to get $4$ , $5$ ways to get $6$ , $5$ ways to get $8$ , and $1$ way to get $12$ . So, $P(B) = \frac{(3+5+5+1)}{(6\times6)}$ .
Simplify Probability B: Simplify $P(B)$ . $P(B) = \frac{14}{36} = \frac{7}{18}$ .