STA 2023 EXAM-2 Practice Problemsvmudunur.myweb.usf.edu/SPRING 2016 Exam-2 Review Solutions.pdf · STA 2023 EXAM-2 Practice Problems Mudunuru, Venkateswara Rao STA 2023 Spring 2016

STA 2023 EXAM-2

Practice Problems

Mudunuru, Venkateswara Rao

STA 2023

Spring 2016

From Chapters 4, 5, & Partly 6 With SOLUTIONS

© Ven M

udun

uru

© Venkateswara Rao Mudunuru Spring 2016 Exam-2 PRACTICE SHEET SOLUTIONS 1

1. A committee of 5 persons is to be formed from 6 men and 4 women. What is the

probability that the committee consists of exactly 2 women?

𝐹𝑜𝑟 𝑛 = 10 𝑎𝑛𝑑 𝑟 = 5 𝑤𝑒 ℎ𝑎𝑣𝑒

𝑛(5 𝑚𝑒𝑚𝑏𝑒𝑟 𝑐𝑜𝑚𝑚𝑖𝑡𝑡𝑒𝑒) = 𝑛(𝑆) = (105

) 𝑜𝑟 10𝐶5 =10!

5! (10 − 5)!= 252 𝑤𝑎𝑦𝑠

Number of committees consisting of exactly 2 women (so the rest of the five must

be 3 men) are, 𝑛(2𝑊 ∩ 3𝑀) = 4𝐶2 × 6𝐶3 = 120 𝑤𝑎𝑦𝑠

𝑃(2𝑊 ∩ 3𝑀) =𝑛(2𝑊 ∩ 3𝑀)

𝑛(5 𝑚𝑒𝑚𝑏𝑒𝑟 𝑐𝑜𝑚𝑚𝑖𝑡𝑡𝑒𝑒𝑠)=

120

252= 0.4762

2. A coin is tossed 5 times. The probability of heads on any toss is 0.57. Let x denote the

number of tails that comes up.

a. Find the probability that tail shows up exactly twice. [Ans: 0.3424] NOTE: If probability of HEADS IS 0.57, PROBABILITY OF tails IS 0.43. Here in the questions, the

success we are looking for is tails showing up. Hence 𝑝 = 0.43

𝐻𝑒𝑟𝑒 𝑟 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑢𝑐𝑐𝑒𝑠𝑠𝑒𝑠 𝑜𝑢𝑡 𝑜𝑓 𝑛 𝑡𝑟𝑎𝑖𝑙𝑠 = 2;

𝑛 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑟𝑖𝑎𝑙𝑠 = 5;

𝑝 = 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑡𝑎𝑖𝑙 𝑜𝑛 𝑒𝑎𝑐ℎ 𝑡𝑟𝑎𝑖𝑙 = 0.43;

𝑞 = 1 − 𝑝 = 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑓𝑎𝑖𝑙𝑢𝑟𝑒 = 1 − 0.43 = 0.57; 𝐻𝑒𝑟𝑒 𝑛 = 5; 𝑟 = 2; 𝑝 = 0.43; 𝑞 = 0.57

𝑃(𝑟) = (𝑛𝑟

) 𝑝𝑟𝑞𝑛−𝑟 =𝑛!

𝑟! (𝑛 − 𝑟)!𝑝𝑟𝑞𝑛−𝑟

𝑃(𝑟 = 2) = 𝑃(2) = (52

) (0.43)2(0.57)5−2 = 0.3424

b. Find the probability that tail shows up less than twice. [Ans: 0.2870]

𝑃(𝑟 < 2) = 𝑃(𝑟 ≤ 1) = 𝑃(0) + 𝑃(1)

𝑈𝑠𝑖𝑛𝑔 𝑃(𝑟) = (𝑛𝑟

) 𝑝𝑟𝑞𝑛−𝑟

= (50

) (0.43)0(0.57)5−0 + (51

) (0.43)1(0.57)5−1

𝑃(𝑟 < 2) = 𝑃(𝑟 ≤ 1) = 0.2870

c. Find the Probability that tail shows up more than twice and less than 5 times.

[Ans: 0.3557]

𝑃(2 < 𝑟 < 5) = 𝑃(3) + 𝑃(4)

= (53

) (0.43)3(0.57)5−3 + (54

) (0.43)4(0.57)5−4 = 0.2583 + 0.0974

𝑃(2 < 𝑟 < 5) = 0.3827 © Ve

n Mud

unuru

http://www.facebook.com/venkatvnv.significantly.insignificant


3. Find the following –

a. In how many different three course meals can be selected from 2 salads, 6 main

dishes, and 4 desserts? [Ans: 48 ways]

𝐴𝑛𝑠𝑤𝑒𝑟: 2 × 6 × 4 = 48 𝑤𝑎𝑦𝑠

b. In how many different ways can I arrange 5 books in a book rack? [Ans: 120]

𝐴𝑛𝑠𝑤𝑒𝑟: 5! = 120 𝑤𝑎𝑦𝑠

c. How many different 5-digit codes are possible without repetition of digits and if the

first digit cannot be zero? [Ans: 27216]

𝐴𝑛𝑠𝑤𝑒𝑟: 9 × 9 × 8 × 7 × 6 = 27216 𝑐𝑜𝑑𝑒𝑠

d. How many different 5-digit codes are possible with repetition of digits and if the

first digit cannot be zero? [Ans: 90000]

𝐴𝑛𝑠𝑤𝑒𝑟: 9 × 10 × 10 × 10 × 10 = 90000 𝑐𝑜𝑑𝑒𝑠

e. How many different 5-digit codes are possible with repetition of digits and if the

first digit can be zero? [Ans: 100000]

𝐴𝑛𝑠𝑤𝑒𝑟: 10 × 10 × 10 × 10 × 10 = 100000 𝑐𝑜𝑑𝑒𝑠

f. Given five women and three men, how many ways can a three-member

committee be formed? [Ans: 56]

𝑛(3 𝑚𝑒𝑚𝑏𝑒𝑟 𝑐𝑜𝑚𝑚𝑖𝑡𝑡𝑒𝑒) = 𝑛(𝑆) = (83

) 𝑜𝑟 8𝐶3 =8!

3! (8 − 3)!= 56 𝑤𝑎𝑦𝑠

4. What is the probability that when we roll a die the result is a number divisible by 1.5, given

that the number is an even number? [Ans: 0.33]

𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛: 𝐻𝑒𝑟𝑒 𝑆 = {1,2,3,4,5,6};

𝐿𝑒𝑡 𝐸𝑣𝑒𝑛𝑡 𝐴 𝑏𝑒 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒 𝑏𝑦 1.5; 𝐴 = {3,6};

𝐸𝑣𝑒𝑛𝑡 𝐵 𝑏𝑒 𝑎𝑛 𝑒𝑣𝑒𝑛 𝑛𝑢𝑚𝑏𝑒𝑟; 𝐵 = {2,4,6}; 𝐴 ∩ 𝐵 = {6}

𝑃(𝐴|𝐵) =𝑃(𝐴 ∩ 𝐵)

𝑃(𝐵)=

16⁄

36⁄

= 0.33 © Ven M

udun

uru



5. A box contains 16 colored mugs: 4 red, 6 yellow and 6 green.

a. A mug is selected at random. What is the probability that the mug chosen is either

red or green? Is this an independent event or dependent event or mutually

exclusive event? [Ans: 0.625]

𝑃(𝑅𝑒𝑑 ∪ 𝐺𝑟𝑒𝑒𝑛) = 𝑃(𝑅𝑒𝑑) + 𝑃(𝐺𝑟𝑒𝑒𝑛) =4

16+

6

16=

10

16= 0.625;

𝑀𝑢𝑡𝑢𝑎𝑙𝑙𝑦𝐸𝑥𝑐𝑙𝑢𝑠𝑖𝑣𝑒

b. A mug is selected at random. What is the probability that the mug chosen is either

red or yellow? [Ans: 0.625]

𝑃(𝑅𝑒𝑑 ∪ 𝑌𝑒𝑙𝑙𝑜𝑤) = 𝑃(𝑅𝑒𝑑) + 𝑃(𝑌𝑒𝑙𝑙𝑜𝑤) =4

16+

6

16=

10

16= 0.625

c. A mug is selected at random. What is the probability that the mug chosen is both

red and yellow? [Ans: 0]

𝑃(𝑅𝑒𝑑 ∩ 𝑌𝑒𝑙𝑙𝑜𝑤) = 0

d. Two mugs are drawn in succession with replacement. What is the probability that

both mugs chosen are yellow? Is this an independent event or dependent event

or mutually exclusive event? [Ans: 0.141]

𝑃(𝑌1 ∩ 𝑌2) = 𝑃(𝑌1) × 𝑃(𝑌2) =6

16×

6

16= 0.141; 𝐼𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡

e. Two mugs are drawn in succession without replacement. What is the probability

that both mugs chosen are yellow? Is this an independent event or dependent

event or mutually exclusive event? [Ans: 0.125]

𝑃(𝑌1 ∩ 𝑌2) = 𝑃(𝑌1) × 𝑃(𝑌2|𝑌1) =6

16×

5

15= 0.125; 𝐷𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡

6. Consider a pack of 52 playing cards. A card is selected at random. Also mention if these

following events are independent or dependent or mutually exclusive?

a. What is the probability that the card is either a diamond or an ace? [Ans: 0.308]

𝑃(𝐷 ∪ 𝐴) = 𝑃(𝐷) + 𝑃(𝐴) − 𝑃(𝐷 ∩ 𝐴) =13

52+

4

52−

1

52= 0.308;

𝑁𝑜𝑡 𝑀𝑢𝑡𝑢𝑎𝑙𝑙𝑦 𝐸𝑥𝑐𝑙𝑢𝑠𝑖𝑣𝑒

b. What is the probability that the card is both a diamond and a king? [Ans: 0.019]

𝑃(𝐷 ∩ 𝐾) =1

52= 0.019

c. What is the probability that the card is either a diamond or a spade? [Ans: 0.5]

𝑃(𝐷 ∪ 𝑆) = 𝑃(𝐷) + 𝑃(𝑆) =13

52+

13

52= 0.5; 𝑀𝑢𝑡𝑢𝑎𝑙𝑙𝑦 𝐸𝑥𝑐𝑙𝑢𝑠𝑖𝑣𝑒 © Ve

n Mud

unuru



d. What is the probability that the card is either a diamond or a face card?[Ans:

0.423]

𝑃(𝐷 ∪ 𝐹) = 𝑃(𝐷) + 𝑃(𝐹) − 𝑃(𝐷 ∩ 𝐹) =13

52+

12

52−

3

52= 0.423;

𝑁𝑜𝑡 𝑀𝑢𝑡𝑢𝑎𝑙𝑙𝑦 𝐸𝑥𝑐𝑙𝑢𝑠𝑖𝑣𝑒

e. What is the probability that the card is both spade and a heart card? [Ans: 0]

𝑃(𝑆 ∩ 𝐻) = 0;

7. A box contains 10 defective and 18 non-defective bulbs. Two are drawn out together.

One of them is tested and found to be non-defective bulb. What is the probability that

the other one is also non-defective? [Ans: 0.63]

𝑃(𝐷2|𝐷1) =𝑃(𝐷1 ∩ 𝐷2)

𝑃(𝐷1)=

1828⁄ × 17

27⁄

1828⁄

= 0.63;

8. A student takes two core courses, statistics and biology. In any given semester the

probability of failing statistics course is 2% and the probability of failing biology is 6%. In a

given semester, what is the probability that

a. Student will fail both the exams? [Ans: 0.0012]

𝑃(𝑆′ ∩ 𝐵′) = 𝑃(𝑆′) × 𝑃(𝐵′) = 0.02 × 0.06 = 0.0012;b. Student will fail one or the other (but not both) courses. [Ans: 0.078]

𝑃((𝑆′ ∩ 𝐵)𝑜𝑟(𝑆 ∩ 𝐵′)) = 𝑃((𝑆′ ∩ 𝐵) ∪ (𝑆 ∩ 𝐵′))

= 𝑃(𝑆′) × 𝑃(𝐵) + 𝑃(𝑆) × 𝑃(𝐵′) = (0.02 × 0.94) + (0.98 × 0.06) = 0.078;

9. Answer the following -

a. If the odds against an event occurring are 5:19, what is the probability of the event

occurring? [Ans: 0.7916]

𝑊𝑒 𝑘𝑛𝑜𝑤 𝑡ℎ𝑎𝑡 𝑂𝑑𝑑𝑠(�̅�) = 𝑛(�̅�): 𝑛(𝐴)

𝑖. 𝑒. , 𝑂𝑑𝑑𝑠(𝑎𝑔𝑎𝑖𝑛𝑠𝑡) = (#𝑎𝑔𝑎𝑖𝑛𝑠𝑡): (#𝑓𝑎𝑣𝑜𝑟𝑖𝑛𝑔) 𝑎𝑛𝑑 𝑛(𝑆) = 𝑛(𝐴) + 𝑛(�̅�)

⇒ 𝑂𝑑𝑑𝑠(𝑎𝑔𝑎𝑖𝑛𝑠𝑡) = 5: 19 𝑖𝑚𝑝𝑙𝑦𝑖𝑛𝑔 𝑡ℎ𝑎𝑡

𝑛(𝑆) = 5 + 19 = 24.

𝑁𝑜𝑤 𝑃(𝐹𝑎𝑣𝑜𝑟𝑖𝑛𝑔) = 𝑛(𝑓𝑎𝑣𝑜𝑟𝑖𝑛𝑔)

𝑛(𝑆)=

19

24= 0.7916

b. Given n (E) = 3, n(S) = 7. What is P (E’)? [Ans: 0.5715]

𝑊𝑒 𝑘𝑛𝑜𝑤 𝑡ℎ𝑎𝑡, 𝑛(𝑆) = 𝑛(𝐸) + 𝑛(𝐸′)

⇒ 7 = 3 + 𝑛(𝐸′) ⇒ 𝑛(𝐸′) = 4.

𝑁𝑜𝑤 𝑃(𝐸′) = 𝑛(𝐸′)

𝑛(𝑆)= 𝑃(𝐸′) =

4

7= 0.5715 © Ve

n Mudun

uru



c. Given n (E) = 3, n(S) = 7. What is Odds (E)? [Ans: 3:4]

𝐻𝑒𝑟𝑒 𝑎𝑔𝑎𝑖𝑛 𝑛(𝑆) = 𝑛(𝐸) + 𝑛(𝐸′) ⇒ 7 = 3 + 𝑛(𝐸′) ⇒ 𝑛(𝐸′) = 4.

𝐻𝑒𝑛𝑐𝑒, 𝑂𝑑𝑑𝑠(𝐸) = 𝑛(𝐸): 𝑛(𝐸′) = 3: 4

d. Given n(E) = x, n(S) = x+y. What is P (E’)? [Ans: y/(x+y)]

𝑈𝑠𝑖𝑛𝑔 𝑛(𝑆) = 𝑛(𝐸) + 𝑛(𝐸′) ⇒ 𝑥 + 𝑦 = 𝑥 + 𝑛(𝐸′) ⇒ 𝑛(𝐸′) = 𝑦.

𝑁𝑜𝑤 𝑃(𝐸′) = 𝑛(𝐸′)

𝑛(𝑆)=

𝑦

𝑥 + 𝑦

e. Given n(E) = x, n(S) = x+y. What are Odds (E’)? [Ans: y:x]

𝑈𝑠𝑖𝑛𝑔 𝑛(𝑆) = 𝑛(𝐸) + 𝑛(𝐸′) ⇒ 𝑥 + 𝑦 = 𝑥 + 𝑛(𝐸′) ⇒ 𝑛(𝐸′) = 𝑦.

𝑂𝑑𝑑𝑠(𝐸)′ = 𝑛(𝐸′): 𝑛(𝐸′) = 𝑦: 𝑥

10. The number of girls and boys enrolled for different majors in a community college are

listed in the table below

Major Girls Boys Total

Biology 150 120 270

Chemistry 100 100 200

Stats 110 130 240

Medicine 175 150 325

Total 535 500 1035

a. If a student is chosen at random, what is the probability that the student is a boy?

[Ans: 0.4830]

𝑃(𝐵𝑜𝑦) =500

1035= 0.4830

b. If a student is chosen at random, what is the probability that the student is a boy

enrolled in stats? [Ans: 0.1256]

𝑃(𝐵𝑜𝑦 ∩ 𝑆𝑡𝑎𝑡𝑠) =130

1035= 0.1256

c. What is the probability that the student is a girl given that she is biology major?

[Ans: 0.5555]

𝑃(𝐺𝑖𝑟𝑙|𝐵𝑖𝑜) =𝑃(𝐺𝑖𝑟𝑙 ∩ 𝐵𝑖𝑜)

𝑃(𝐵𝑖𝑜)=

1501035⁄

2701035⁄

= 0.5555 © Ven M

udun

uru



d. What is the probability that the student is biology major given that she is a girl?

[Ans: 0.2804]

𝑃(𝐵𝑖𝑜|𝐺𝑖𝑟𝑙) =𝑃(𝐺𝑖𝑟𝑙 ∩ 𝐵𝑖𝑜)

𝑃(𝐺𝑖𝑟𝑙)=

1501035⁄

5351035⁄

= 0.2804

11. Below is a survey of 615 students regarding satisfaction with the choice their majors.

Maths Biology Physics English Total Very satisfied 27 47 13 12 99 Satisfied 61 39 52 30 182 Not Satisfied 58 99 95 82 334 Total 146 185 160 124 615

Assume that the sample represents the entire population of students. Find the probability

that a student selected at random is

a. At least satisfied with her/his major. [Ans: 0.4569]

𝑃(𝐴𝑡 𝑙𝑒𝑎𝑠𝑡 𝑆𝑎𝑡𝑖𝑠𝑓𝑖𝑒𝑑) =99 + 182

615= 0.4569

b. Satisfied, given that she/he is a Mathematics major. [Ans: 0.4178]

𝑃(𝑆𝑎𝑡𝑖𝑠𝑓𝑖𝑒𝑑|𝑀𝑎𝑡ℎ) =𝑃(𝑆𝑎𝑡𝑖𝑠𝑓𝑖𝑒𝑑 ∩ 𝑀𝑎𝑡ℎ)

𝑃(𝑀𝑎𝑡ℎ)=

61615⁄

146615⁄

= 0.4178

c. Not satisfied, given that she/he is a Biology major. [Ans: 0.5351]

𝑃(𝑁𝑜𝑡 𝑆𝑎𝑡𝑖𝑠𝑓𝑖𝑒𝑑|𝐵𝑖𝑜) =𝑃(𝑁𝑜𝑡 𝑆𝑎𝑡𝑖𝑠𝑓𝑖𝑒𝑑 ∩ 𝐵𝑖𝑜)

𝑃(𝐵𝑖𝑜)=

99615⁄

185615⁄

= 0.5351

d. A Physics major. [Ans: 0.2602]

𝑃(𝑃ℎ𝑦) =160

615= 0.2602

e. Not an English major. [Ans: 0.7984]

𝑃(𝑁𝑜𝑡 𝐸𝑛𝑔) = 1 − 𝑃(𝐸𝑛𝑔) = 1 −124

615= 0.7984

f. Satisfied and Not Satisfied. [Ans: 0]

𝑃(𝑆𝑎𝑡𝑖𝑠𝑓𝑖𝑒𝑑 ∩ 𝑁𝑜𝑡 𝑆𝑎𝑡𝑖𝑠𝑓𝑖𝑒𝑑) =0

615= 0

© Ven M

udun

uru



12. A card is drawn from a deck of 52 cards.

a. What is the probability that the card is diamond or a face card? [Ans: 0.4231]

𝑃(𝐹 ∪ 𝐷) = 𝑃(𝐹) + 𝑃(𝐷) − 𝑃(𝐹 ∩ 𝐷) =12

52+

13

52−

3

52= 0.4231

b. What is the probability that the card is not a spade card? [Ans: 0.75]

𝑃(𝑆′) = 1 −13

52= 0.75

c. What is the probability that the card is either a 6 or heart card? [Ans: 0.3077]

𝑃(6 ∪ ℎ𝑒𝑎𝑟𝑡) = 𝑃(6) + 𝑃(ℎ𝑒𝑎𝑟𝑡) − 𝑃(6 ∩ ℎ𝑒𝑎𝑟𝑡) =4

52+

13

52−

1

52= 0.3077

d. What is the probability that the card is a king or queen or jack? [Ans: 0.2308]

𝑃(𝐾 ∪ 𝑄 ∪ 𝐽) = 𝑃(𝐾) + 𝑃(𝑄) + 𝑃(𝐽) =4

52+

4

52+

4

52= 0.2308

e. What are the odds in favor of selecting a face card? [Ans: 3:10]

𝑂𝑑𝑑𝑠(𝑓𝑎𝑐𝑒 𝑐𝑎𝑟𝑑) = (#𝑓𝑎𝑐𝑒 𝑐𝑎𝑟𝑑𝑠): (#𝑛𝑜𝑡 𝑓𝑎𝑐𝑒 𝑐𝑎𝑟𝑑𝑠) = 12: 40 = 3: 10

13. A coin is tossed 5 times. The probability of tails on any toss is 0.52. Let x denote the number

of tails that comes up.

a. Find the probability that tail shows up at least twice. [Ans: 0.8365]



𝐻𝑒𝑟𝑒: 𝑛 = 5; 𝑝 = 0.52; 𝑞 = 1 − 𝑝 = 1 − 0.52 = 0.48; 𝑎𝑛𝑑 𝑟 = 2;

𝑃(𝑟 ≥ 2) = 1 − 𝑃(𝑟 ≤ 1) = 1 − [𝑃(0) + 𝑃(1)]

= 1 − [5𝐶0(0.52)0(0.48)5−0 + 5𝐶1(0.52)1(0.48)5−1] = 1 − [0.1635] = 0.8365

b. Find the probability that tail shows up in all tosses. [Ans: 0.0380]

𝑃(𝑟 = 5) = 5𝐶5(0.52)5(0.48)5−5 = 0.038

c. Find the Probability that tail shows up more than twice and less than 5 times.

[Ans: 0.4994]

𝑃(2 < 𝑟 < 5) = 𝑃(3) + 𝑃(4) = [5𝐶3(0.52)3(0.48)5−3 + 5𝐶4(0.52)4(0.48)5−4] = 0.4994 © Ven M

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14. The average number of injuries per week in an industry are recorded as 2.1.

a. Find the probability that in a given week there can be no injuries? [Ans: 0.1224]

𝐺𝑖𝑣𝑒𝑛 𝑎𝑣𝑒𝑟𝑎𝑔𝑒, 𝜆 = 2.1; 𝑊𝑒 𝑤𝑖𝑙𝑙 𝑢𝑠𝑒 𝑃𝑜𝑖𝑠𝑠𝑜𝑛 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛.

𝑃(𝑟) =𝑒−𝜆𝜆𝑟

𝑟!

𝑃(𝑟 = 0) =𝑒−2.1 × (2.1)0

0!= 0.1224

b. Find the probability that in a given week, there will be less than 3 injuries?

[Ans: 0.6496]

𝑃(𝑟 < 3) = 𝑃(𝑟 ≤ 2) = 𝑃(0) + 𝑃(1) + 𝑃(2)

=𝑒−2.1 × (2.1)0

0!+

𝑒−2.1 × (2.1)1

1!+

𝑒−2.1 × (2.1)2

2!= 0.6496

c. Find the probability that in a given week, there will be at least 3 injuries? [Ans: 0.3504]

𝑃(𝑟 ≥ 3) = 1 − 𝑃(𝑟 ≤ 2) = 1 − [𝑃(0) + 𝑃(1) + 𝑃(2)]

𝑃(𝑟 ≥ 3) = 1 − 0.6496 = 0.3504 {Using result of part b, above}

d. Find the probability that in a given week, there will be at least 2 but at most 4

injuries? [Ans: 0.5582]

𝑃(2 ≤ 𝑟 ≤ 4) = 𝑃(2) + 𝑃(3) + 𝑃(4)

=𝑒−2.1 × (2.1)2

2!+

𝑒−2.1 × (2.1)3

3!+

𝑒−2.1 × (2.1)4

4!= 0.5582

e. What is the mean and standard deviation of this problem? [Ans: 2.1; 1.449]

𝑀𝑒𝑎𝑛 𝜇 = 𝜆 = 2.1;

𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 𝜎 = √𝜆 = √2.1 = 1.449

© Ven M

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15. An insurance company has determined that on the average it receives 3 accident

claims per day.

𝐺𝑖𝑣𝑒𝑛 𝑎𝑣𝑒𝑟𝑎𝑔𝑒, 𝜆 = 3; 𝑊𝑒 𝑤𝑖𝑙𝑙 𝑢𝑠𝑒 𝑃𝑜𝑖𝑠𝑠𝑜𝑛 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛.

𝑃(𝑟) =𝑒−𝜆𝜆𝑟

𝑟!

a. Find the probability that the company receives at least 4 accident claims on a

randomly selected day. [Ans: 0.3528]

𝑃(𝑟 ≥ 4) = 1 − 𝑃(𝑟 ≤ 3) = 1 − [𝑃(0) + 𝑃(1) + 𝑃(2) + 𝑃(3)]

= 1 − [𝑒−3 × (3)0

0!+

𝑒−3 × (3)1

1!+

𝑒−3 × (3)2

2!+

𝑒−3 × (3)2

3!] = 0.3528

b. Find the probability that the company receives no claims on a randomly selected

day. [Ans: 0.0498]

𝑃(𝑟 = 0) =𝑒−3 × (3)0

0!= 0.0498

16. Assuming that the heights of boys in a high-school basketball tournament are normally

distributed with mean 80 inches and standard deviation 2 inches

a. How many boys in a group of 40 are expected to be taller than 84 inches?

[Ans: 1]

𝐶𝑙𝑒𝑎𝑟𝑙𝑦, 𝑤𝑒 ℎ𝑎𝑣𝑒 2.5% 𝑜𝑓 40 𝑏𝑜𝑦𝑠 𝑤ℎ𝑜 𝑎𝑟𝑒 𝑡𝑎𝑙𝑙𝑒𝑟 𝑡ℎ𝑎𝑛 84 height.

Hence 2.5

100×40=1⟹There is 1 Boy in a group of 40 who is taller than 84" height.

© Ven M

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Start

3,6

1

2

3

4

5

6

1,2,4,5H

T

P(3)=P(6)=1/6

P(1)=P(2)=P(4)=P(5)=1/6

P(1)=…=P(6)=1/6

0.5

b. What percentage of boys falls between 78 and 84 inches height?

[Ans: 0.815 or 81.5%]

𝑇ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 68% 𝑜𝑓 𝑏𝑜𝑦𝑠 𝑤ℎ𝑜 𝑎𝑟𝑒 𝑓𝑎𝑙𝑙𝑖𝑛𝑔 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 78 𝑡𝑜 82 & 13.5% 𝑓𝑟𝑜𝑚 82 𝑡𝑜 84, 𝑠𝑜,

𝑇ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 𝑎 𝑡𝑜𝑡𝑎𝑙 𝑜𝑓 68% + 13.5% = 81.5% 𝑜𝑓 𝑏𝑜𝑦𝑠

𝑤ℎ𝑜 𝑎𝑟𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 78 𝑎𝑛𝑑 84 𝑖𝑛𝑐ℎ𝑒𝑠

c. What percentage of boys falls below 82 inches? [Ans: 84%]

𝑇ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 50% 𝑜𝑓 𝑏𝑜𝑦𝑠 𝑤ℎ𝑜 𝑎𝑟𝑒 𝑓𝑎𝑙𝑙𝑖𝑛𝑔 𝑏𝑒𝑙𝑜𝑤 80 𝑖𝑛𝑐ℎ𝑒𝑠 & 34% 𝑓𝑟𝑜𝑚 80 𝑡𝑜 82 𝑖𝑛𝑐ℎ𝑒𝑠. 𝑠𝑜

𝑇ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 𝑎 𝑡𝑜𝑡𝑎𝑙 𝑜𝑓 50% + 34% = 84% 𝑜𝑓 𝑏𝑜𝑦𝑠 𝑤ℎ𝑜 𝑎𝑟𝑒 𝑏𝑒𝑙𝑜𝑤 82 𝑖𝑛𝑐ℎ𝑒𝑠

17. Given the experiment: a fair die is rolled; if a multiple of three appears uppermost, then

the die is rolled again, otherwise a fair coin is tossed.

a. Represent this scenario using a tree diagram with the respective probabilities on

each branch mentioned. [Ans: ]

b. What is the sample space of this experiment?

𝑆 = {(3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6),

(1, 𝐻), (1, 𝑇), (2, 𝐻), (2, 𝑇), (4, 𝐻), (4, 𝑇), (5, 𝐻), (5, 𝑇)}

c. What is the associated probability distribution in this experiment?

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

= (6,6) = 0.02778

(1, 𝐻) = (1, 𝑇) = (2, 𝐻) = (2, 𝑇) = (4, 𝐻) = (4, 𝑇) = (5, 𝐻) = (5, 𝑇) = 0.08333

© Ven M

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18. Given the experiment: a fair die is rolled followed by tossing a fair coin.

a. Represent this scenario using a tree diagram with the respective probabilities on

each branch mentioned. [Ans: try or guess ]

b. What is the sample space of this experiment?

𝐴𝑛𝑠: 𝑆 = {1𝐻, 1𝑇, 2𝐻, 2𝑇, … 6𝐻, 6𝑇}

c. What is the associated probability distribution in this experiment?

𝐴𝑛𝑠: 1𝐻 = (1/6) ∗ (1/2) = 0.0833 = 1𝑇 = 2𝐻 = 2𝑇 = … = 6𝐻 = 6𝑇

19. A TV company when purchasing thousands electronic components apply this sampling

plan: randomly select 15 of them and then accept the whole batch if there are at most

3 defective.

a. If a particular lot has a 3% of defective components, what is the probability that

the whole lot is accepted? [Ans: 0.9991]

Solution: 𝑛 = 15; 𝑝 = 0.03; 𝑞 = 1 − 𝑝 = 1 − 0.03 = 0.97; 𝑎𝑛𝑑

Using:



𝑃(𝑟 ≤ 3) = 𝑃(0) + 𝑃(1) + 𝑃(2) + 𝑃(3)

= 15𝐶0(0.03)0(0.97)15−0 + 15𝐶1(0.03)1(0.97)15−1 + 15𝐶2(0.03)2(0.97)15−2

+ 15𝐶3(0.03)3(0.97)15−3 = 0.9991

b. In a sample of 2000 electronic components, what will be the expected number of

the defective components and what is the standard deviation?

[Ans: 60; 7.6288]

Also since given 𝑎 𝑠𝑎𝑚𝑝𝑙𝑒 𝑜𝑓 𝑠𝑖𝑧𝑒 𝑛 = 2000; 𝑎𝑛𝑑 𝑝 = 0.03;

𝑇ℎ𝑒 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑑𝑒𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡𝑠 𝑎𝑟𝑒

𝐸(𝑋) = 𝜇 = 𝑛𝑝 = 2000 × 0.03 = 60

𝑇ℎ𝑒 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦 𝜎 = √𝑛𝑝𝑞

𝜎 = √2000 × 0.03 × 0.97 = 7.6288 © Ven M

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Start

HeadHead

Tail

TailHead

Tail

20. Draw a probabilistic tree diagram for the following experiment: a fair coin is tossed; if a

head appears uppermost, then the fair coin is tossed again, otherwise an unfair coin with

P (Head) = 0.55 is tossed. What is the sample space of this experiment? What is the

associated probability distribution in this experiment?

𝑆𝑎𝑚𝑝𝑙𝑒 𝑠𝑝𝑎𝑐𝑒, 𝑆 = {𝐻𝐻, 𝐻𝑇, 𝑇𝐻, 𝑇𝑇}

𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝐷𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛:

𝐻𝐻 = 𝐻𝑇 = 0.5 × 0.5 = 0.25;

𝑇𝐻 = 0.5 × 0.55 = 0.275;

𝑇𝑇 = 0.5 × 0.45 = 0.225

21. Consider the following experiment. One card is selected from a deck of cards. Find the

probability the card selected is:

a. A six number card b. A seven & a king card c. A diamond or a club card

d. A king card e. A five & a diamond card f. A face or an ace card

g. A red card h. A ten or an ace card i. A face or a heart card

j. A four and a black card k. A queen and a king card l. A red and a black card

© Ven M

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a. A six number card

𝑃(6) =4

52= 0.0769

b. A seven & a king card

𝑃(7 ∩ 𝐾) =0

52= 0

c. A diamond or a club card

𝑃(𝐷 ∪ 𝐶) =13

52+

13

52= 0.5

d. A king card

𝑃(𝐾) =4

52= 0.0769

e. A five & a diamond card

𝑃(5 ∩ 𝐷) =1

52= 0.0192

f. A face or an ace card

𝑃 (𝐹 ∪ 𝐴) = 12

52+

4

52= 0.3077

g. A red card

𝑃(𝑅) = 26

52= 0.5

h. A ten or an ace card

𝑃(10 ∪ 𝐴) = 4

52+

4

52= 0.1538

i. A face or a heart card

𝑃(𝐹 ∪ 𝐻) = 12

52+

13

52−

3

52= 0.4231

j. A four & a black card

𝑃(4 ∩ 𝐵) = 2

52= 0.0384

k. A queen and a king card

𝑃(𝑄 ∩ 𝐾) = 0

52= 0

l. A red and a black card

𝑃(𝑅 ∩ 𝐵) = 0

52= 0

22. A quality control engineer is in charge of testing whether or not 90% of the DVD players

produced by his company conform to specifications. To do this, the engineer randomly

selects a batch of 12 DVD players from each day’s production. The day’s production is

acceptable provided no more than 1 DVD player fails to meet specifications. Otherwise,

the entire day’s production has to be tested. What is the probability that the engineer

incorrectly passes a day’s production as acceptable if only 80% of the day’s DVD players

actually conform to specification?

𝑛 = 12; 𝑝 = 0.2; 𝑞 = 1 − 𝑝 = 1 − 0.2 = 0.8; 𝑎𝑛𝑑

Using:



𝑃(𝑟 ≤ 1) = 𝑃(0) + 𝑃(1)

= 12𝐶0(0.2)0(0.8)12−0 + 12𝐶1(0.2)1(0.8)12−1 = 0.275

23. Are the following experiments follow binomial distribution or not?

Give your answer with Yes/No followed by (fixed number of trials, possible outcomes).

a. Surveying 84 people to determine if they like dessert after dinner.

Yes (84, like or dislike)

b. Rolling a die 70 times to find a prime number occur.

Yes (70, Prime or Not a prime) c. Drawing a card from a deck and getting a Jack card

Yes (1, jack or not a jack) d. Surveying 100 people which cuisine they prefer the most.

No (100, more than 2 possible cuisines) e. Testing 8 different brands of phones to see which brand is best to purchase

No (NA, 8 brands) © Ven M

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f. Testing 1 brand of Tylenol by using 10 people to determine whether it is effective

Yes (10, effective or not) g. Surveying 143 high school students to see what would they prefer as their major for

a college degree duties in a row.

No (143, more than 2 majors) 24. A die is rolled. Find the

𝐻𝑒𝑟𝑒 𝑙𝑒𝑡 𝑆 = 𝑠𝑒𝑡 𝑜𝑓 𝑎𝑙𝑙 𝑝𝑜𝑠𝑖𝑠𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑤ℎ𝑒𝑛 𝑤𝑒 𝑟𝑜𝑙𝑙 𝑎 𝑑𝑖𝑒.

𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝐸 = {1,2,3,4,5,6} ⇒ 𝑛(𝑆) = 6

a. Odds in favor of getting a number less than 2 [Ans: 1:5]

𝐻𝑒𝑟𝑒 𝑙𝑒𝑡 𝐸 = 𝑠𝑒𝑡 𝑜𝑓 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 2. 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝐸 = {1} ⇒ 𝑛(𝐸) = 1

𝑎𝑛𝑑 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, 𝑛(𝐸′) = 𝑛(𝑆) − 𝑛(𝐸) = 6 − 1 = 5.

𝑂𝑑𝑑𝑠(𝐸) = 𝑛(𝐸): 𝑛(𝐸′) = 1: 5

b. Odds in favor of getting a number less than 4 [Ans: 1:1]

𝐻𝑒𝑟𝑒 𝑙𝑒𝑡 𝐸 = 𝑠𝑒𝑡 𝑜𝑓 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 4. 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝐸 = {1,2,3} ⇒ 𝑛(𝐸) = 3


𝑂𝑑𝑑𝑠(𝐸) = 𝑛(𝐸): 𝑛(𝐸′) = 3: 3 = 1 ∶ 1

c. Odds against getting a number more than 5 [Ans: 5:1

𝐻𝑒𝑟𝑒 𝑙𝑒𝑡 𝐸 = 𝑠𝑒𝑡 𝑜𝑓 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑚𝑜𝑟𝑒 𝑡ℎ𝑎𝑛 2. 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝐸 = {1} ⇒ 𝑛(𝐸) = 1


𝑂𝑑𝑑𝑠(𝐸′) = 𝑛(𝐸′): 𝑛(𝐸) = 5: 1

d. Probability of getting an even prime [Ans: 0.1667]

𝐻𝑒𝑟𝑒 𝑙𝑒𝑡 𝐸 = 𝑠𝑒𝑡 𝑜𝑓 𝑒𝑣𝑒𝑛 𝑝𝑟𝑖𝑚𝑒 𝑛𝑢𝑚𝑏𝑒𝑟𝑠. 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝐸 = {2} ⇒ 𝑛(𝐸) = 1

𝑃(𝐸) =1

6= 0.1667

e. Probability of getting a multiple of 4 [Ans: 0.1667]

𝐻𝑒𝑟𝑒 𝑙𝑒𝑡 𝐸 = 𝑠𝑒𝑡 𝑜𝑓 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑒𝑠 𝑜𝑓 4. 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝐸 = {4} ⇒ 𝑛(𝐸) = 1

𝑃(𝐸) =1

6= 0.1667

f. Probability of getting a number divisible by 1.75 [Ans: ]

𝐻𝑒𝑟𝑒 𝑙𝑒𝑡 𝐸 = 𝑠𝑒𝑡 𝑜𝑓 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒 𝑏𝑦 1.75. 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝐸 = { } ⇒ 𝑛(𝐸) = 0

𝑃(𝐸) =0

6= 0 © Ve

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More for Exam-2:

1) The probability of the sample space is ONE.

2) The probability of any event is between 0 and 1

3) Given 𝑷(𝑬) = 𝟎. 𝟑𝟖, then 𝑷(𝑬𝑪) = 𝟎. 𝟔𝟐

4) Let 𝒏(𝑬) be the number of outcomes in an event 𝐸. Given 𝒏(𝑬) = 𝟐𝟏 and 𝑷(𝑬) =

𝟎. 𝟕, then 𝒏(𝑬𝑪) = 𝟗

5) Given 𝒏(𝑬) = 𝟐𝟏 and 𝑷(𝑬) = 𝟎. 𝟕, then 𝑶(𝑬) = 𝟐𝟏: 𝟗 𝒐𝒓 𝟕: 𝟑.

6) Given 𝑨 and 𝑩 are dependent then 𝑷(𝑨 ∩ 𝑩) = 𝑷(𝑨)𝑷(𝑩|𝑨) 𝒐𝒓 𝑷(𝑩)𝑷(𝑨|𝑩)

7) Given 𝑨 and 𝑩 are independent then 𝑷(𝑨|𝑩) = 𝑷(𝑨)

8) Given 𝑨 and 𝑩 are independent then 𝑷(𝑨 ∩ 𝑩) = 𝑷(𝑨)𝑷(𝑩)

9) Given 𝑨 and 𝑩 are mutually exclusive then 𝑷(𝑨 ∩ 𝑩) = 𝟎.

10) Given 𝑨 and 𝑩 are mutually exclusive then 𝑷(𝑨|𝑩) = 𝟎

11) Given 𝑨 and 𝑩 are mutually exclusive then 𝑨 ∩ 𝑩 = 𝝓

12) Given two independent random variables with 𝑬(𝒙𝟏) = 𝝁𝟏, 𝑽𝒂𝒓(𝒙𝟏) = 𝝈𝟏𝟐, 𝑬(𝒙𝟐) =

𝝁𝟐, and 𝑽𝒂𝒓(𝒙𝟐) = 𝝈𝟐𝟐; if we transform the data using the linear combination 𝒚 =

𝟑𝒙𝟏 − 𝟓𝒙𝟐, then 𝑬(𝒚) = 𝟑𝝁𝟏 − 𝟓𝝁𝟐 and 𝑽𝒂𝒓(𝒚) = 𝟗𝝈𝟏𝟐 + 𝟐𝟓𝝈𝟐

𝟐

13) In a normal distribution, the empirical rule states approximately 68% of the

observed data fall within one standard deviation of the mean.

14) In a normal distribution, the empirical rule states approximately 95% of the

observed data fall within two standard deviation of the mean.

15) In a normal distribution, the empirical rule states approximately 99.7% of the

observed data fall within three standard deviation of the mean.

16) Given two independent random variables with 𝑬(𝒙𝟏) = 𝟓, 𝑽𝒂𝒓(𝒙𝟏) = 𝟒, 𝑬(𝒙𝟐) = 𝟔,

and 𝑽𝒂𝒓(𝒙𝟐) = 𝟗; if we transform the data using the linear combination 𝒚 = 𝒙𝟏 −

𝟒𝒙𝟐, then 𝑬(𝒚) = −𝟏𝟗 and 𝑽𝒂𝒓(𝒚) = 𝟏𝟒𝟖

17) Given two independent random variables with 𝑬(𝒙𝟏) = 𝟓, 𝑽𝒂𝒓(𝒙𝟏) = 𝟏𝟔,; if we

transform the data using the linear combination 𝒚 = 𝟐 − 𝟑𝒙𝟏, then

𝑬(𝒚) = −𝟏𝟑 and 𝑽𝒂𝒓(𝒚) = 𝟏𝟒𝟒

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Helpful Rules:

𝐸𝑥𝑎𝑐𝑡𝑙𝑦 𝑘 𝑠𝑢𝑐𝑐𝑒𝑠𝑠𝑒𝑠

𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 𝑜𝑟 𝑛𝑜 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 𝑘 𝑠𝑢𝑐𝑐𝑒𝑠𝑠𝑒𝑠

𝑎𝑡 𝑚𝑜𝑠𝑡 𝑜𝑟 𝑛𝑜 𝑚𝑜𝑟𝑒 𝑡ℎ𝑎𝑛 𝑘 𝑠𝑢𝑐𝑐𝑒𝑠𝑠𝑒𝑠

𝐿𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 𝑘 𝑠𝑢𝑐𝑐𝑒𝑠𝑠𝑒𝑠

𝑀𝑜𝑟𝑒 𝑡ℎ𝑎𝑛 𝑘 𝑠𝑢𝑐𝑐𝑒𝑠𝑠𝑒𝑠

𝑟 = 𝑘 𝑟 ≥ 𝑘 𝑟 ≤ 𝑘 𝑟 < 𝑘 𝑟 > 𝑘

𝐶𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑎𝑙 𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑃(𝐴|𝐵) =𝑃(𝐴 ∩ 𝐵)

𝑃(𝐵)

𝑂𝑑𝑑𝑠(𝐸) 𝑂(𝐸) = 𝑛(𝐸): 𝑛(𝐸′) 𝑂𝑑𝑑𝑠(𝐸′) 𝑂(𝐸′) = 𝑛(𝐸′): 𝑛(𝐸)

𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑎𝑛 𝑒𝑣𝑒𝑛𝑡 𝑃(𝐸) = 𝑛(𝐸)

𝑛(𝑆)

𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑎 𝑐𝑜𝑚𝑝𝑙𝑖𝑚𝑒𝑛𝑡 𝑃(𝐸′) = 𝑛(𝐸′)

𝑛(𝑆) 𝑜𝑟 1 − 𝑃(𝐸)

𝐹𝑎𝑐𝑡𝑜𝑟𝑖𝑎𝑙 𝑛! = 𝑛 × (𝑛 − 1) × (𝑛 − 2) … × 2 × 1

𝐶𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛𝑠 (𝑛𝑟

) 𝑜𝑟 𝑛𝐶𝑟 =𝑛!

𝑟! (𝑛 − 𝑟)!

𝑃𝑒𝑟𝑚𝑢𝑡𝑎𝑡𝑖𝑜𝑛𝑠 𝑛𝑃𝑟 =𝑛!

(𝑛 − 𝑟)!

𝐵𝑖𝑛𝑜𝑚𝑖𝑎𝑙 𝑃(𝑟) = 𝑛𝐶𝑟 × 𝑝𝑟 × 𝑞𝑛−𝑟; 𝑞 = 1 − 𝑝

𝑃𝑜𝑖𝑠𝑠𝑜𝑛 𝑃(𝑟) =𝑒−𝜆𝜆𝑟

𝑟!

𝑛(𝑆) = 𝑛(𝐸) + 𝑛(𝐸′) 𝑛(𝐸′) = 𝑛(𝑆) − 𝑛(𝐸)

𝑛(𝑆) = 1

𝐹𝑒𝑤 𝑞𝑢𝑖𝑐𝑘 𝑒𝑥𝑎𝑚𝑝𝑙𝑒𝑠 𝑜𝑓 𝑆𝑎𝑚𝑝𝑙𝑒 𝑆𝑝𝑎𝑐𝑒𝑠

𝑇𝑜𝑠𝑠𝑖𝑛𝑔 𝑎 𝑐𝑜𝑖𝑛 𝑆 = {𝐻, 𝑇}

𝑇𝑜𝑠𝑠𝑖𝑛𝑔 𝑡𝑤𝑜 𝑐𝑜𝑖𝑛𝑠 𝑆 = {𝐻𝐻, 𝐻𝑇, 𝑇𝐻, 𝑇𝑇}

𝑇𝑜𝑠𝑠𝑖𝑛𝑔 𝑡ℎ𝑟𝑒𝑒 𝑐𝑜𝑖𝑛𝑠 𝑆 = {𝐻𝐻𝐻, 𝐻𝐻𝑇, 𝐻𝑇𝐻, 𝐻𝑇𝑇, 𝑇𝐻𝐻, 𝑇𝐻𝑇, 𝑇𝑇𝐻, 𝑇𝑇𝑇}

𝑅𝑜𝑙𝑙𝑖𝑛𝑔 𝑎 𝑑𝑖𝑒 𝑆 = {1,2,3,4,5,6} 𝑅𝑜𝑙𝑙𝑖𝑛𝑔 𝑡𝑤𝑜 𝑑𝑖𝑐𝑒 𝑆 = {(1,1), (1,2), (1,3) … , (6,4), (6,5), (6,6)}

𝑅𝑢𝑙𝑒 𝐸𝑥𝑎𝑚𝑝𝑙𝑒 𝐸(𝑎) = 𝑎 𝐸(2) = 2

𝐸(𝑎𝑋) = 𝑎𝐸(𝑋) 𝐸(−5𝑋) = −5𝐸(𝑋)

𝑉(𝑎) = 0 𝑉(4) = 0

𝑉(𝑎𝑋) = 𝑎2𝑉(𝑋) 𝑉(−3𝑋) = (−3)2𝑉(𝑋) = 9𝑉(𝑋)

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© Ven M

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STA 2023 EXAM-2 Practice Problemsvmudunur.myweb.usf.edu/SPRING 2016 Exam-2 Review Solutions.pdf · STA 2023 EXAM-2 Practice Problems Mudunuru, Venkateswara Rao STA 2023 Spring 2016

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