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Solved 17 Questions of Probability and Statistics - Review Exam 2 | MATH 1530, Exams of Probability and Statistics

Material Type: Exam; Professor: Maxson; Class: Probability and Statistics; Subject: Mathematics; University: Walters State Community College; Term: Unknown 1989;

Typology: Exams

Pre 2010

Uploaded on 08/19/2009

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Math 1530 Walters State Comm College Review for Exam 2
1. In rolling a die, what is the probability of obtaining either a 5 or a 6?
1 1 1
(5 6) (5) (6)
6 6 3
P P P
= + = + =
2. On four consecutive rolls of a die a 6 is obtained. What is the probability of obtaining a 6 on the 5
th
roll?
Events are independent …
1
(6)
6
P
=
3. Based on meteorological records, the probability that it will snow in a certain town on Jan 17 is 0.315. Find the probability
that it will not snow in that particular town on Jan 17.
The probability of the compliment of an event is: 1 – probability of the event = 1 – 0.315 = 0.685
4. In tossing a coin, what is the probability of obtaining HHHTH in that order in five tosses?
Events are independent …
5
1 1
( ) ( ) ( ) ( ) ( ) ( )
2 32
P H H H T H P H P H P H P T P H
= = =
5. Consider a standard deck of 52 cards. Determine the following probabilities.
A. P(9 of clubs): 1/52
B. P(ace): 4/52 = 1/13
C. P(ace of spades and ace of hearts) in that order with and without replacement
: ; without:
52 52 2704 52 51 2652
with = =
D. P(3 aces) with and without replacement
1 1 1 1 1 1 1 1
: ; without:
13 13 13 2197 13 17 25 5525
with = =
E. P(ace, king, queen) in that order with and without replacement
3
1 1 1 4 2 8
: ; without: 0.000483
13 2197 13 51 25 16575
with = =
6. An urn contains 20 red balls, 20 green balls, and 20 yellow balls. Suppose three balls are randomly selected. Find the
following probabilities.
A. P(3 red) with and without replacement
20 20 20 1 1 19 9 57
: ( ) ; without:
60 60 60 27 3 59 29 1711
with P red red red = = =
pf3
pf4

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Download Solved 17 Questions of Probability and Statistics - Review Exam 2 | MATH 1530 and more Exams Probability and Statistics in PDF only on Docsity!

Math 1530 Walters State Comm College Review for Exam 2

  1. In rolling a die, what is the probability of obtaining either a 5 or a 6?

P ∪ = P + P = + =

  1. On four consecutive rolls of a die a 6 is obtained. What is the probability of obtaining a 6 on the 5

th roll?

Events are independent …

P =

  1. Based on meteorological records, the probability that it will snow in a certain town on Jan 17 is 0.315. Find the probability

that it will not snow in that particular town on Jan 17.

The probability of the compliment of an event is: 1 – probability of the event = 1 – 0.315 = 0.

  1. In tossing a coin, what is the probability of obtaining HHHTH in that order in five tosses?

Events are independent …

5

1 1

( ) ( ) ( ) ( ) ( ) ( )

P H H H T H P H P H P H P T P H

  1. Consider a standard deck of 52 cards. Determine the following probabilities.

A. P(9 of clubs): 1/

B. P(ace): 4/52 = 1/

C. P(ace of spades and ace of hearts) in that order with and without replacement

: ; without:

with ⋅ = ⋅ =

D. P(3 aces) with and without replacement

: ; without:

with ⋅ ⋅ = ⋅ ⋅ =

E. P(ace, king, queen) in that order with and without replacement

3

1 1 1 4 2 8

: ; without: 0.

with

  1. An urn contains 20 red balls, 20 green balls, and 20 yellow balls. Suppose three balls are randomly selected. Find the

following probabilities.

A. P(3 red) with and without replacement

: ( ) ; without: with P redredred = ⋅ ⋅ = ⋅ ⋅ =

B. P(3 balls of any one color) without replacement

C. P(3 different colors) with replacement

1 or.... 3!

  1. A coin is tossed 500 times with the results being 240 heads and 260 tails. What is the relative frequency (empirical)

probability of getting a head from this experiment?

  1. Two dice, one red and one white are tossed. Determine the probability of:

A. P(2 white dice): an impossible event … 0

B. P(7): 6/36 = 1/

C. P(red = 6): 1/

D. P(red = 6 | sum = 11): only 2 ways to get sum of 11, red 6 white 5 or white 6 red 5 … 1/

  1. A red die and a white die are tossed. Consider these events to find the following probabilities.

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. P(M): 6/36 = 1/

B. P(N): 1/

C. P(M or N): 6/36 + 1/36 = 7/

D. P(M and N): it is impossible to have a red die = 4 and the sum be 12 … P(M and N) = 0

E. P(O): 18/36 = 1/

F. P(M | P): 3/18 = 1/

M. Red die = 4

N. sum is 12

O. sum is even

P. white die = even

Sample space for 2 die

Answers: (Probabilities may be written as decimals rounded to 3 significant digits, reduced fractions, or percents. I’ve used

different methods in the answers. Any of the 3 ways is considered correct.

  1. 1/3 2. 1/6 3. 0.685 4. (1/2)

5 = 1/

5a. 1/52 5b. 1/13 5c. with: 1/2704 5c. w/o: 1/

5d. with: 1/2197 5d. w/o: 1/5525 5e. with: 1/2197 5e.w/o: 0.

6a. with: 1/27 6a. w/o: 0.0333 6b. 0.102 6c 0.

  1. 240/500 = 0.48 8a. 0 8b. 1/6 8c. 1/

8d. ½ 9a. 1/6 9b. 1/36 9c. 7/

9d. 0 9e. ½ 9f. 3/18 = 1/6 10a. 71/

10b. 37/513 10c. 605/1058 10d. 0.35976 or 0.360 10e. 37/

10f. 74/545 13a. 1 – (.1)

8 = 1 13b. (.9)

8 = 0.430 14. 2,600,

  1. 13! = 6,227,020,800 16. 20 C 4 = 4,845 17. 20 P 4 = 116,