Two number cubes are rolled for two separate events:

all of the combinations of numbers on both cubes that give a sum less than 10.
all combinations of numbers on both cubes that give a sum that is a multiple of 3.

a. in terms of a reduced fraction, find the conditional probability of b given that a occurs first. fill in the blanks p( __ l __ ) = __ / __

b. complete the conditional-probability formula for event b given that event a occurs first by writing a and b in the blanks: p( __ l __ ) = p( __ ∩ __ ) / p( __ )

c. list the members of the sample space for event a ∩ b.

d. list the members of the sample space for event a.

Step-by-step explanation:

When two cubes are rolled then the favourable outcomes are: 36 (given below)

(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)

Favourable events of numbers on both cubes that give a sum less than 10 are (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);(5,1) ; (5,2) ; (5,3) ;(5,4) ;(6,1) ; (6,2) ; (6,3)

so total no. of  favourable events of numbers on both cubes that give a sum less than 10 = 30

Probability of getting sum less than 10 =  

Favourable events of numbers on both cubes that give a sum that is a multiple of 3 are (1,3);(1,6);(2,3);(2,6);(3,1) ; (3,2) ; (3,3) ;(3,4) ; (3,5) ;(3,6);(4,3);(4,6);(5,3);(5,6);(6,1) ; (6,2) ; (6,3) ;(6,4) ; (6,5) ;(6,6)

so total no. of  favourable events of numbers on both cubes that give a sum that is a multiple of 3 = 20

Probability of getting a multiple of 3=

(A) P(B|A)= \frac{P(B∩A)}{P(A)} = = =0.5  

(B)P(A|B)== =0.75  

(C) {A∩B} = {3, 6, 9, 12, 15, 18}

(D) {A} = {1, 2, 3, 4, 5 ,6, 7, 8, 9}l

They are on each side of the number line. -9 is 9 units left of zero and 9 is 9 units right of zero

Step-by-step explanation: