Statistics Fall2013 - Tutorial Material - c2p2 - Oct12 - Added Solution

8/13/2019 Statistics Fall2013 - Tutorial Material - c2p2 - Oct12 - Added Solution

http://slidepdf.com/reader/full/statistics-fall2013-tutorial-material-c2p2-oct12-added-solution 1/5

VIETNAM NATIONAL UNIVERSITY – HCMC STATISTICS FOR BUSINESSINTERNATIONAL UNIVERSITY Academic Year 2013 – 2014, Semester 01

of 5

DATE: SATURDAY, OCTOBER 12, 2013

TUTORIAL MATERIAL

---o0o---

ADDITIONAL PROBLEM 01:

Assume we have a bag containing 4 green balls and 6 red balls. A person randomly selects one

of the balls and then without replacement of the first ball randomly selects a second ball. What is

the probability of selecting 2 green balls?

ANSWER: CONDITIONAL PROBABILITY

Let A be the event that the 1st ball is green

B be the event that 2nd

ball is green

( ) =4 10⁄ ,() = 3 9⁄ . Hence, ( ∩ ) = (4 10⁄ )(3 9⁄ ) = = 0.1333


There are about 80% of people in this city play football. 90% of people go to the stadium for

watching a football match at weekend. There are 75% also play football and watch football

match at weekend.

a. What is the percentage of people dislike football (neither playing football nor watch a football

match)?

b. Given that a chosen people watch football match, what is the probability that people also play

football?

c. Given that a chosen people do not watch football match, what is the probability that people

play football?

d. Given that a chosen people watch football match, what is the probability that people do not

play football?




of 5


ANSWER: (BASIC RULES OF PROBABILITY)

ProbabilityPlaying football

()

Not playing football

() Total

Watching football

match() 0.75 0.15 0.90

Not watching

football match

( )

0.05 0.05 0.10

Total 0.80 0.20 1.00

(a).

( ∩ ) = 0.05

(b).

(|) = ( ∩ )

() =

0.75

0.90= 0.8333

(c).

(| ) =( ∩ )

( ) =

0.05

0.10= 0.5

(d).

(|) =( ∩ )

() =

0.15

0.90= 0.1667


Whenever a certain NBA basket player goes to the foul line for two shots, he makes his first shot

with probability 0.75. If he makes the first shot, then he makes the second shot with probability

0.80; if he misses the first shot, then he makes the second shot with probability 0.70. What is the

probability that he makes only one of the shots?

ANSWER: CONDITIONAL PROBABILITY

Let A be the event of making the 1st shot

B be the event of making the 2nd

shot

Making the 1st shot( ) = .

Making the 2nd

shot (| ) = 0.80

Missing the 2nd

shot (| ) = 0.20

Missing the 1st

shot( ) = . Making the 2

nd

shot (| ̅) = 0.70 Missing the 2

nd shot (| )̅ = 0.30

The probability that the player makes the 1st shot and misses the 2

nd shot is:

( ∩ ) = ( )(| ) = (0.75)(0.20) = 0.15




of 5


The probability that the player misses the 1st shot and makes the 2

nd shot is:

( ̅ ∩ ) = ( )̅(| )̅ = (0.25)(0.70) = 0.175

Hence, the probability that the player makes only one of the shots is:

( ∩ ) + ( ̅ ∩ ) = 0.15 + 0.175 = 0.325


An urn contains 5 red, 6 blue, and 9 green balls. If a set of 3 green balls is randomly selected,

what is the probability that each of the balls will be

(a) of the same color.

(b) of different colors.

ANSWER: (COMBINATIONAL CONCEPTS)

(a). The probability that each of balls will be of the same color is

(53) + (63) + (93)

(203) =

1

10= 0.1

(b). The probability that each of balls will be of the different color is

(51) ×(61) ×(91)(203)

= 938

= 0.2368


Consider 3 boxes. Box A contains 4 red and 5 blue balls; box B contains 3 red and 7 blue balls,

and box C contains 6 reds and 6 blue balls.

(a). If a ball is randomly selected, what is the probability that it is a blue ball?

(b). If a ball is randomly selected, what is the probability that it is from box B given that it is a

red ball?




of 5


ANSWER: (THE LAW OF TOTAL PROBABILITY AND BAYES’S THEOREM)

Box A

(1/3)

Red ball (4/9)

Blue ball (5/9)

Box B (1/3) Red ball (3/10)Blue ball (7/10)

Box C

(1/3)

Red ball (6/12)

Blue ball (6/12)

(a).

( ) = (1/ 3)(5/ 9) + (7/ 10) + (6/ 12) = 79

135= 0.5852

(b).

( | =1

3 3

10

13

49

+ 310

+ 612

=

27

112= 0.2411


A card is drawn from a standard deck of 52 cards and discarded (i.e. not replaced). A secondcard is drawn from the remaining deck of 51 cards.

(a). What is the probability that the second card is a spade?

(b). Given that the second card was a spade, what is the probability that the first card was also a

spade?

ANSWER: (THE LAW OF TOTAL PROBABILITY AND BAYES’S THEOREM)

The 1st card is a spade

(13/52)

The 2nd

card is a spade (12/51)

The 2n

card is not a spade (39/51)

The 1st card is not a spade

(39/52)

The 2n

card is a spade (13/51)

The 2nd

card is not a spade (38/51)




of 5


(a).

(ℎ 2 ) = (13/ 52)(12/ 51) + (39/ 52)(13/ 51) = 0.25

(b).

(ℎ 1 |ℎ 2 )

=(13/ 52)(12/ 51)

(13/ 52)(12/ 51) + (39/ 52)(13/ 51) =

4

17= 0.2353

Statistics Fall2013 - Tutorial Material - c2p2 - Oct12 - Added Solution

Documents

Transcript of Statistics Fall2013 - Tutorial Material - c2p2 - Oct12 - Added Solution