R Exercise 1.

(a) Simulate the rolling of two dice 10,000 times.
(b) Identify which rolls of the dice are in the event A, the dice add up to a perfect square (4 or 9). Determine what proportion of the 10,000 rolls are in A.
(c) Identify which rolls of the dice are in the event B, the dice add up to an even number. Determine what proportion of the 10,000 rolls are in B.
(d) Find out which rolls are in A ∩ B. Find the proportion that are in A ∩ B. How does that compare to the proportion in A multiplied by the proportion that are in B?
(e) Of the rolls in which B occurs that you identified in part (c), what proportion of those rolls are also in A. How does this compare to the proportion you computed in part (b)?
(f) Do your results in this R exercise match what you would expect based on the answers you got in problem (4) of the previous sectio

p ( A ) = 7/36

p ( B ) = 1/2

p ( A ∩ B ) = 1/12

p ( A / B ) = 1/6

p ( B / A ) = 3/7

Step-by-step explanation:

Given:

- Two dices are rolled together n = 10,000 of times

Find:

a) Identify which rolls of the dice are in the event A, the dice add up to a perfect square (4 or 9).

Determine what proportion of the 10,000 rolls are in A.

Solution:

- Identity all outcomes per roll such that the sum is a perfect square i.e 4 or 9.

  Sum is 4 or 9: ( 1 , 3 ) , ( 2 , 2 ) , ( 3 , 1 ) , ( 3 , 6 ) , ( 5 , 4 ) , ( 4 , 5 ) , ( 6 , 3 )

  Possible outcomes per roll = 7.

  Total Outcomes of sums per roll = 36

- Since each roll is independent from each other the proportion of 10,000 rolls in event A would be:

                  Event ( A ) = n* ( Possible Outcomes / Total Outcomes )

                  Event ( A ) = 10,000* ( 7 / 36 ) = 1944.44

                  Proportion p(A) = Event (A) / n = 1944.44 / 10,000 = 0.1944

Find:

b) Identify which rolls of the dice are in the event B, the dice add up to an even number.

Determine what proportion of the 10,000 rolls are in B.

Solution:

- Identity all outcomes per roll such that the sum is even 2 , 4 , 6 , 8 , 10 , 12.

  Sum is even: ( 1 , 1 ) , ( 1 , 3 ) , ( 1 , 5 ) , ( 2 , 2 ) , ( 2 , 4 ) , ( 2 , 6 ) ,  ( 3 , 1 ) , ( 3 , 3 ) , ( 3 , 5 ) , ( 4 , 2 ) , ( 4 , 4 ) , ( 4 , 6 ) ( 5 , 1 ) , ( 5 , 3 ) , ( 5 , 5 ) , ( 6 , 2 ) , ( 6 , 4 ) , ( 6 , 6 )

  Possible outcomes per roll = 18.

  Total Outcomes of sums per roll = 36

- Since each roll is independent from each other the proportion of 10,000 rolls in event B would be:

                  Event ( B ) = n* ( Possible Outcomes / Total Outcomes )

                  Event ( B ) = 10,000* ( 18 / 36 ) = 5,000

                  Proportion p(B) = Event (B) / n = 5,000 / 10,000 = 0.5

Find:

c) Find out which rolls are in A ∩ B. Find the proportion that are in A ∩ B. How does that compare to the proportion in A multiplied by the proportion that are in B?

- The events outcomes common in both events are as follows:

   A ∩ B = ( 1 , 3 ) , ( 2 , 2 ) , ( 3 , 1 )

   Possible outcomes per roll = 3.

   Total Outcomes of sums per roll = 36

- Since each roll is independent from each other the proportion of 10,000 rolls in event A ∩ B would be:

                  Event ( A ∩ B ) = n* ( Possible Outcomes / Total Outcomes )

                  Event ( A ∩ B ) = 10,000* ( 3 / 36 ) = 833.33

                  Proportion p(A ∩ B) = Event (A ∩ B) / n = 833.333 / 10,000 = 1/12

                  p(A) * p(B) = 0.5*0.1944 = 0.09722 > 1/12

Find:

d) Of the rolls in which B occurs what proportion of those rolls are also in A. How does this compare to the proportion you computed in part (b)?

- Possible outcomes common in both A and B = 3

- p ( A / B ) = p ( A ∩ B ) / p(B)

                   = ( 1 / 12 ) / 0.5

                   = 1 / 6

- p ( B / A ) = p ( A ∩ B ) / p(A)

                   = ( 1 / 12 ) / (7/36)

                   = 3 / 7

6mm^3 or .6cm^3

      216

      /   \

  36     6

  /   \     /

6     6   6

factor trees are nice.

6x = 48

i hope this . 6 cards each and a total of $48. dvide each side by 6 and you will have your answer.