Question

Consider the experiment of rolling a pair of dice. Suppose that we are interested in the sum of the face values showing on the dice. (a) How many sample points are possible? (Hint: use the counting rule for multiple-step experiments.) 36 (b) List the sample points. There to sum the face values of a pair of dice to 2. There to sum the face values of a pair of dice to 3. There to sum the face values of a pair of dice to 4. There to sum the face values of a pair of dice to 5. There to sum the face values of a pair of dice to 6. There to sum the face values of a pair of dice to 7. There to sum the face values of a pair of dice to 8. There to sum the face values of a pair of dice to 9. There to sum the face values of a pair of dice to 10. There to sum the face values of a pair of dice to 11. There to sum the face values of a pair of dice to 12. (c) What is the probability of obtaining a value of 5? (d) What is the probability of obtaining a value of 8 or greater? (e) Because each roll has six possible even values (2, 4, 6, 8, 10, and 12) and only five possible odd values (3, 5, 7, 9, and 11), the dice should show even values more often than odd values. Do you agree with this statement? Explain. This statement correct because P(odd) = and P(even) = . (f) What method did you use to assign the probabilities requested? classical method empirical method subjective method relative frequency method

Answer 1

Answer:

Check Explanation.

Step-by-step explanation:

a) Sample points refer to the individual possible outcomes in a probability experiment.

For this experiment, there are 36 sample points for the roll of two dice.

b) The sample points for each sum value include

2

(1,1)

3

(1,2) (2,1)

4

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

5

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

6

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

7

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

8

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

9

(3,6) (4,5) (5,4) (6,3)

10

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

11

(5,6) (6,5)

12

(6,6)

c) Probability of obtaining a sum value of 5

There are 4 sample points that give 5,

Hence, probability of getting a 5 = (4/36) = (1/9) = 0.1111

d) Probability of obtaining a sum value of 8 or greater = (number of sample points that give a sum value of 8, 9, 10, 11 and 12)/(total number of sample points)

Probability of obtaining a sum value of 8 or greater = (5+4+3+2+1)/36 = (15/36) = (5/12) = 0.4167

e) Let's check with the actual probabilities.

P(odd) = probability of obtaining an odd number sum value = (number of sample points that give an odd number sum value)/total sum value

P(odd) = (number of sample points that give 3,5,7,9 and 11)/36

P(odd) = (2+4+6+4+2)/36 = (18/36) = (1/2) = 0.5

P(even) = 1 - P(odd) = 0.5

The reasoning that each roll has six possible even values (2, 4, 6, 8, 10, and 12) and only five possible odd values (3, 5, 7, 9, and 11), the dice should show even values more often than odd values is wrong because it is the sample points (sum of numbers) that actually determine how often an even or an odd number turns up, not the actual odd number or even number.

And the sample points (sum of numbers) that give even numbers (18) are just as much as the ones that give odd numbers (18 too).

f) The method used in obtaining the probability of each sample point is the classical method of assigning probabilities.

In classical method, each possible outcome has an equal chance of occurring and an experiment with n simple outcomes assigns a probability of (1/n) to each outcome.

But the method of assigning probabilities to each sum of values is closer to the relative frequency approach.

Answer 2

Answer:

Step-by-step explanation:

a) the sample point of the random experiment is 36

b) (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)

Only (1,1) will sum to give 2.

(1,2) and (2,1) will sum up to give 3.

(1,3), (2,2) (3,1) will sum up to give 4.

(1,4) (2,3) (3,2) (4,1) will sum up to give 5.

(1,5) (2,4) (3,3) (4,2) (5,1) will sum up to give 6

(1,6) (2,5) (3,4) (4,3) (5,2) (6,1) will sum up to give 7

(2,6) (3,5) (4,4) (5,3) (6,2) will sum up to give 8

(3,6) (4,5) (5,4) (6,3) will sum up to give 9

(4,6) (5,5) (6,4) will sum up to give 10

(5,6) (6,5) will sum up to give 11

(6,6) 1ill sum up to give 12.

c) probability of obtaining a value of 5 = (number of possible times 5 will occur)/ (total outcome) = $\frac{4}{36} =\frac{1}{9}$

d) probability of obtaining a value of 8 or greater = $\frac{15}{36} = \frac{5}{12}$

e) yes, since the total number is 11 and probability of obtaining odd number is $\frac{5}{11}$ and probability of obtaining even number is $\frac{6}{11}$.

P(odd)$\neq$P(even)

f) emprical method was used