Question

13. - 18 points ASWSBE13 4.E.016, My Notes Ask Your Teacher 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. (w) How many sample points are possible? (Hint: use the counting rule for multiple-step experiments.) (b) List the sample points. There Select to sum the face values of a pair of dice to 2. There-Sec to sum the face values of a pair of dice to 3. There Select to sum the face values of a pair of dice to 4. There-Select to sum the face values of a pair of dice to 5. There--Select to sum the face values of a pair of dice to 6. There--Select- to sum the face values of a pair of dice to 7. There ---Select to sum the face values of a pair of dice to 8. There--Select to sum the face values of a pair of dice to 9. There--Select to sum the face values of a pair of dice to 10. There ---Select to sum the face values of a pair of dice to 11. There ---Select to sum the face values of a pair of dice to 12. (c) What is the probability of obtaining a value of 7? (d) What is the probability of obtaining a value of 10 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 -Select- correct because plod) - and Pleven) . What method did you use to assign the probabilities requested? classical method subjective method

Answer 1

a-

As there are 6 faces in a dice so there will be- 6*6=36 sample points.

b-

Following are the sample points to the numbers from 2-12:

2-Only one (1,1)

3-Two sample points (2,1) (1,2)

4-Three sample points (3,1)(1,3)(2,2)

5-Four sample points (1,4)(4,1)(2,3)(3,2)

6-Five sample points(1,5)(5,1)(2,4)(4,2)(3,3)

7-Six sample points (1,6)(6,1)(2,5)(5,2)(3,4)(4,3)

8-Five sample points (6,2)(2,6)(3,5)(5,3)(4,4)

9-Four sample points(3,6)(6,3)(4,5)(5,4)

10-Three sample points (5,5)(6,4)(4,6)

11-Two sample points (6,5)(5,6)

12-One sample points ( 6,6)

c-

P(7)= (sample points of 7/ total sample space)

P(7)= 6/36= 1/6

d-

P(value of 10 or greater)= 6/36=1/6

e-

This statement is correct as P(odd)=5/11 and the P(even)=6/11 as sum as 1 is not possible so the number of odds are less than even.

f-

As the events are equally likely as we do not consider it a biased coin so we are applying the classical method here.