Question

An experiment consists of casting a pair of dice and observing the number that falls uppermost on each die.

Answer 1

Answer:

a) $\text{Probability}=\frac{7}{12}$

b) $\text{Probability}=\frac{1}{9}$

Step-by-step explanation:

Given : An experiment consists of casting a pair of dice and observing the number that falls uppermost on each die.

When two dice rolled once the outcome will be,

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)  

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)  

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)  

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)  

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)  

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)  

(a) Determine the event that the sum of the numbers falling uppermost is less than or equal to 7.

Solution :

$\text{Probability}=\frac{\text{Favorable outcome}}{\text{Total number of outcome}}$

Favorable outcome are (the sum of the numbers falling uppermost is less than or equal to 7)

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5)

(3,1) (3,2) (3,3) (3,4)

(4,1) (4,2) (4,3)

(5,1) (5,2)

(6,1)

So, Favorable outcome = 21

Total number of outcome = 36

$\text{Probability}=\frac{21}{36}$

$\text{Probability}=\frac{7}{12}$

(b) Determine the event that the number falling uppermost on one die is a 4 and the number falling uppermost on the other die is greater than 4.

Solution :

$\text{Probability}=\frac{\text{Favorable outcome}}{\text{Total number of outcome}}$

Favorable outcome are (the number falling uppermost on one die is a 4 and the number falling uppermost on the other die is greater than 4.)

(4,5) (4,6) or (5,4) (6,4)

So, Favorable outcome = 4

Total number of outcome = 36

$\text{Probability}=\frac{4}{36}$

$\text{Probability}=\frac{1}{9}$    

Answer 2

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

b.)
(4,5) (4,6) or (5,4) (6,4)