1271829760_NSS_M1V2_12_FS__eng_

12 Some Special Discrete Probability Distributions


Classwork

Classwork (p. 12.4)

(a) Yes (d) Yes

(b) No (e) No

(c) No (f) Yes

Quick Practice

Quick Practice 12.1 (p. 12.5)(a) S follows a Bernoulli distribution with p = 0.88.

(i) s 0 1P(S = s) 0.12 0.88

(ii)

(b)

Quick Practice 12.2 (p. 12.6)

(a)

(b)

Quick Practice 12.3 (p. 12.9)(a)

(b) Each throw is a Bernoulli trial (the dice is either thrown to

obtain a “6” with probability or a number other than “6”

with probability ), and each throw is independent

of each other.

∴ X ~ B(5, ).

∴

(c) (i)

(ii)

(iii)

(d)

Quick Practice 12.4 (p. 12.11)Let N be the random variable representing the number of private

cars crossing the Western Harbour Tunnel, then N ~ B(7, ).

(a)

(b)

(c)

(d)

Quick Practice 12.5 (p. 12.11)Let X and Y be the random variables representing the number of heads obtained from coin A and B respectively, then X ~ B(3, 0.5) and Y ~ B(4, 0.5),i.e.

and

.

NSS Mathematics in Action Module 1 Vol.2 Full Solutions

Quick Practice 12.6 (p. 12.12)Let X be the random variable representing the number of defective light bulbs found in the sample, then X ~ B(30, 3%).i.e.

.

∴ It is higher than the previous sampling plan.

Quick Practice 12.7 (p. 12.14)(a) Let X and Y be the random variables representing the number

of defective iPods found in Plan A and B respectively, then X ~ B(20, 0.05) and Y ~ B(10, 0.05),i.e.

and

.

∴ Plan B has a higher probability of acceptance for thebatch.

(b) Let M and N be the random variables representing the number of defective iPods found in Plan A and B respectively, then M ~ B(20, 0.1) and N ~ B(10, 0.1),i.e.

and

.

∴ Plan B has a higher probability of acceptance for the batch.

Quick Practice 12.8 (p. 12.21)Each throw is a Bernoulli trial (the dice is either thrown to obtain a

“6” with probability or a number other than “6” with probability

), and each throw is independent of each other.

∴ X ~ Geo .

∴

(a)

(b)

(c)

Quick Practice 12.9 (p. 12.22)Let random variable X be the number of dishes required until Mary can get the Toro sushi, then X ~ Geo (0.15).i.e.

(a)

58


(b)

(c)

Quick Practice 12.10 (p. 12.24)Let random variable X be the number of purchases until Mrs. Leung can get the letter ‘S’, then X ~ Geo (0.1).

i.e.

(a)

(b)

(c) E(X)

∴ The expected amount for Mrs. Leung to get an “S”

∴ In terms of the money paid, she should buy the letter “S” from Mrs. Ho.

Quick Practice 12.11 (p. 12.25)Let random variable N be the number of times until the player can hit the target, then N ~ Geo (0.3).

(a)

(b)

(c) (cor. to 4 sig. fig.)

(d) (cor. to 4 sig. fig.)

Quick Practice 12.12 (p. 12.31)X ~ Po(3)

i.e. , where x = 0, 1, 2, …

(a)

(b)

(c)

Quick Practice 12.13 (p. 12.33)As there are about 6000 WBC per mm3 blood on average, then

there are on average WBC in a drop of 0.001 mm3

blood.Let X be the random variable representing the number of WBC in a drop of 0.001 mm3 blood, then X ~ Po(6),

i.e. , where x = 0, 1, 2, …

Quick Practice 12.14 (p. 12.34)Let X and Y be the random variables representing the number of tea and coffee sold per minute respectively, then X follows a Poisson

distribution with mean and Y follows a Poisson distribution

with mean .

i.e. , where x = 0, 1, 2, … and

, where y = 0, 1, 2, …

(a) (i)

(ii)


(iii)

(iv)

(b)

Quick Practice 12.15 (p. 12.40)(a) Let X be the random variable representing the number of

tosses, then X ~ B(5, 0.65).i.e.

P (X = 3)

(b) Let Y be the random variable representing the number of tosses required until the first heads is gotten,then Y ~ Geo (0.65).i.e.


60


(a) Let X be the random variable representing the number of Scandium nuclei undergo decay, then X ~ B(8, 0.8).i.e.

(b) Let Y be the random variable representing the minute in which the Scandium nucleus first undergoes decay, then Y ~ Geo (0.8).i.e.

(c) Let Z be the random variable representing the number of Scandium nuclei will not undergo decay in the coming 4 minutes but will undergo decay in the 5th minute, then Z ~ B(8, 0.00128).i.e.

Quick Practice 12.17 (p. 12.43)(a) X ~ B(3, 0.6)

i.e.

(b)

(c) Y ~ B(5, 0.648)i.e.

(d)


(a) M ~ Po(1.2), i.e. , where x =

0, 1, …(i)

(ii)

(b) Let Y be the random variable representing the number of pages with fewer than 3 misprints, then Y ~ B(20, 0.879487098)i.e.

(i)

(ii)

(iii) Let Z be the random variable representing the number of pages with no misprints, then Z ~ B(20, 0.301194211).

(iv)

Exercise

Exercise 12A (p. 12.14)Level 1

1. (a) X follows a Bernoulli distribution with .

(b)

(c)

2. (a)

(b)


3. (a)

(b) (i) x 0 1

(ii)

(iii)

(c)

4. (a)

(b)

(c)

(d)

5. (a) X follows the binomial distribution where X ~ B(10, 0.3).

(b)

6. (a)

(b) Let A be the random variable representing the amount donated.

7. Let X be the random variable representing the number of games team A loses, then X ~ B(6, 0.4).i.e.

P (X = 2)

8. Let W be the event that Ma Lin wins the match.P (Ma Lin is the champion)

9. Let X be the random variable representing the number of

questions answered correctly, then X ~ B(10, ).

i.e.

If the student passes the test, he must correctly answer 6 questions.

10. Let X be the random variable representing the number of staff drive their cars to work, then X ~ B(14, 0.7).

62


i.e.

11. Let X be the random variable representing the number of patients recover after receiving the treatment, then X ~ B(8, 0.24).i.e.

(a) P(X = 8)

(b) P(X 6)

(c) P(X < 3)

12. Let N be the random variable representing the number of components meet the specifications, then N ~ B(50, 0.9).

,

13. Let X be the random variable representing the number of defective components, then X ~ B(6, 0.15)i.e.

(a) (i) P(X = 0)

(ii) P (X = 1)

(iii) P (X < 4)

(iv)

(b) ,


children having brown hair, then X ~ B .

(a)

(b)

(c)

Level 2

15. (a)

(b)

16. (a) S follows the Bernoulli distribution.

(b)

(c)

(d)


(e) ∵ t = 1 or 4 ∴ T does not have the same kind of probability

distribution as S.

17. Let X and Y be the random variables representing the number of winning home matches and away matches respectively, then X ~ B(3, 0.7) and Y ~ B(5, 0.4),i.e.

and

.

(a)

(b)

(c)

18. P (Draw)

As Tom and Mary have the same chance of winning a game if no draw occurs.

∴ P (Mary wins)

Let X be the random variable representing the number of

games that Mary wins, then X ~ B

P(X = 2) (cor. to 4 sig. fig.)

19. Let X and Y be the random variables representing the number of defective items in stage 1 and 2 respectively, then X ~ B(5, 0.1) and Y ~ B(10, 0.1),

(a) P(X = 0)

(b)

fig.) sig. 4 to(cor. 08657.0

])9.0)(1.0()9.0(1[])9.0)(1.0([

)]1()0(1)[1(

)21(

9101

10451

CC

YPYPXP

YXP

(c)

(d)

20. (a) Let C be the event that a student correctly by guessing.

E(Total mark)

(b) Let K be the event that a student correctly.

21. (a) Let X be the random variable representing the number of days that the stock rises $2, then X ~ B(4, 0.55)

(b) The following table shows the probability distribution of the stock price:

Price 58 54

P(X = x) (0.55)4

Price 50 46

P(X = x)

Price 42

64


P(X = x) (0.45)4

Let S be the price after 4 days.

∴ The expected value and the standard deviation of the stock price after 4 days is $50.8 and $3.980 respectively.

22. Let X be the random variable representing the number of colour blind persons in a sample of size n, then X ~ B(n, 0.02).

∴ The minimum number of people to be selected is 94.

23.

Solving (1) and (2), we have

24.

∴ The minimum value of n is 3.

25.

26.

Let X be the random variable representing the number of “6” obtained, then X ~ B(10, 0.2).(a)

(b)

(c)

27. (a) X ~ B(n, p)

Solving (1) and (2), we have

(b)

28. Let X be the random variable representing the number of days Tom has to wait more than 5 minutes, then X ~ B(5, 0.6). (a) (i)


(ii)

(b)

(c) Sample mean , standard deviation = 1.2∵ Both figures are close to the answers obtained in (b). ∴ Tom’s belief is true.

Exercise 12B (p. 12.26)Level 11. (a)

(b)

(c)

2. (a)

(b)

(c)

(d)

3. (a)

(b)

(c)

(d)

4. Let X be the random variable representing the number of interviews until complete a successful interview, then X ~ Geo (0.28).i.e.

(a)

(b)

(c)

66


5. Let X be the random variable representing the number of tosses until heads is obtained, then X ~ Geo (0.5).i.e.

(a)

(b)

(c)

(d)

6. P (success)

Let X be the random variable representing the number of draws required until the first “success” occurs,

then X ~ Geo .

i.e.

(a) (cor. to 4 sig.

fig.)

(b)

(c)

7. P (Success)

Let X be the random variable representing the number of tosses required until the first “success” occurs,

then X ~ Geo .

i.e.

(a)

(b)

(c) (cor. to 4 sig. fig.)

(d)

… etc∵ P(X = 1) is the largest.∴ The most possible number of trials is 1.

8. Let K be the random variable representing the number of ball

drawn until a blue ball is drawn, then K ~ Geo .

i.e.

(a)


(b)

9. Let X be the random variable representing the number of free-throws until Jordan makes it, then X ~ Geo(0.3).i.e.

(a)

(b)

10. P (the number greater than or equal to ‘2’)

Let X be the random variable representing the number of rolls required until the number obtained is greater than or equal to ‘2’, then

X ~ Geo .

Level 2

11. (a) P(‘2’ is obtained)


rolls until a ‘2’ is obtained, then X ~ Geo .

(b) P(at least one ‘2’ obtained) = 1 – P (no ‘2’ obtained)

Let Y be the random variable representing the number of

rolls until at least one ‘2’ is obtained, then Y ~ Geo

.

(c) P (success) = 1 – P (no ‘2’ obtained)

Let Y be the random variable representing the number of rolls required until at least one ‘2’ is obtained, then

Y ~ Geo .

∴ The minimum number of dice required is 4.

12. (a) Let N be the random variable representing the number of throws required until a win is obtained, then

N ~ Geo .

i.e. P(N = n) , where n = 1, 2, …

The required probability

(b) The required probability

(cor. to 4 sig. fig.)

68


(c)

∴ The required ratio is 6 : 5.

13. ∵ There are total 4 heads occur in 10 tosses and the 9th and 10th tosses are heads.

∴ There are total 2 heads occur in the first 8 tosses.∴ The required probability

14. (a) Let A be the random variable representing the number of days required until station A makes the first correct forecast, then A ~ Geo(0.6).

(b) Let B be the random variable representing the number of days required until station B makes the first correct forecast, then B ~ Geo(0.8).

(c) P(Both stations are correct on a day)

Let X be the random variable representing the number of days required until both stations make the first correct forecast together, then X ~ Geo(0.48).

P (X = 3)

15. (a) P (a red ball is drawn)

Let X be the random variable representing the number of draws required until a red ball is drawn, then

X ~ Geo .

P (X = m)

(b)

16. (a) P(the sponsor will not need to donate)

(b) Let X be the random variable representing the number of shots required until Mr. Wong can score, then X ~ Geo(0.75).i.e.

(c) E(money donated by the sponsor)

17. Let X be the random variable representing the number of cells observed until a mutated cell is observed, then X ~ Geo(p).i.e.


(a)

(b) (i)

(ii)

(iii)

18. Let X be the random variable representing the number of shots required until the target is hit, then X ~ Geo(0.3).i.e.

∴ The largest value of n is 6.

19. (a)

Let X be the random variable representing the number of draws required until a silver coin is drawn, then

X ~ Geo .

(b)

Let Y be the random variable representing the number of draws required after the 1st silver coins is drawn until the

second silver coin is drawn, then Y ~ Geo .

∴

20.

∵ Totally 3 ‘6’s occur in 7 throws and the 7th throw is ‘6’.∴ Totally 2 ‘6’s occur in the first 6 throws.∴ The required probability

21. Let X be the random variable representing the number of forms required until the first form of an applicant from a family with parents who are musicians is screened, then X ~ Geo(0.35).i.e.

(a)

(b) ∵ The 5th enrolment form is the 3rd applicant that comes from a family with parents who are musicians.

∴ There are total 2 forms from a family with parents who are musicians in the first 4 forms.

∴ The required probability

22. Let X be the random variable representing the number of children until the second son is born.i.e.

(a)

(b) Let Y be the random variable representing the number of children required until a boy is born, then Y ~ Geo(0.45).

The mean number of children

70


(c) ∵ There are exactly 2 sons.∴

(d)

…etc∴ P(X = 3) is the largest.∴ The couple will most likely have 3 children.

Exercise 12C (p. 12.35)Level 11. (a)

(b)

2. (a) (cor. to 4 sig.

fig.)

(b)

3. (a)

∴

(b)

4. Let X be the random variable representing the number of customers buying on-sale appliances, then X ~ Po(5).

5. (a) (cor. to 4 sig.

fig.)

(b) P(no complaint in 7 days)

6. As there are on average 10 customers in a 5-minute interval, then there are on average 2 customers in a 1-minute interval.Let X be the random variable representing the number of customers in a 1-minute interval, then X ~ Po(2).

(a) (cor. to 4 sig. fig.)

(b)


(c)

(d)

7. As there are on average 90 customers in one hour, then there are on average 1.5 customers in a 1-minute interval.Let X be the random variable representing the number of customers in a 1-minute interval, then X ~ Po(1.5).

(a) (cor. to 4 sig.

fig.)

(b) P(at least 1 customer enter in 2 min)= 1 – P (no customer enter in 2 min)

8. (a) As there are on average 15 incoming calls customers in a minute, then there are on average 2.5 customers in 10 seconds.Let X be the random variable representing the number of incoming calls in 10 seconds, then X ~ Po(2.5).

(b) As there are on average 15 incoming calls customers in a minute, then there are on average 7.5 customers in 30 seconds.Let Y be the random variable representing the number of incoming calls in 30 seconds, then Y ~ Po(7.5).

9. (a) Let X be the random variable representing the number of amoeba in a 10 mL solution, then X ~ Po(5).(i)

(ii) (cor. to 4 sig.

fig.)(iii)

(b) As there are on average 5 amoeba in a 10 mL solution, then there are on average 10 amoeba in a 20 mL solution.Let Y be the random variable representing the number of amoeba in a 20 mL solution, then Y ~ Po(10).

(i) (cor. to

4 sig. fig.)

(ii)

(c) As there are on average 5 amoeba in a 10 mL solution, then there are on average 2.5 amoeba in a 5 mL solution.Let Z be the random variable representing the number of amoeba in a 20 mL solution, then Z ~ Po(2.5).

(i) (cor. to

4 sig. fig.)(ii)

72


10. (a) Mean

Standard deviation (cor. to 4 sig. fig.)

(b) We have to check if all of the following holds:I. The occurrences are randomly scattered in the

specified interval and they are independent.II. Two or more occurrences cannot occur

simultaneously.III. The expected number of occurrences in any interval

is proportional to the length of the specified interval.Since III seems not hold for the number of goals per match, a Poisson distribution may not be an appropriate distribution. (or any reasonable answer)

11. According to the condition in 10(b), the number of parcels may not follow condition II. Thus a Poisson distribution may not be an appropriate distribution. (or any reasonable answer)

Level 212. Let X be the random variable representing the number of

particles emitted in 1 second, then X ~ .

13. Let X be the random variable representing the number of customer in a 30-minute interval and be the mean of the

distribution, then X ~ .

As the number of the customers in a 30-minute interval is on average 2.995732274, then there are on average

(cor. to 4 sig. fig.) customers in

a 15-minute interval.

∴

14. Let X be the random variable representing the daily demand for digital cameras and be the mean of the distribution, then

X ~ .

(a)

(b) The required probability

15. (a) As there are on average 1.67 cars passing through the tunnel per minute, then there are on average 8.35 cars passing through the tunnel in a 5-minute interval.

Let X be the random variable representing the number of cars passing through the tunnel in a 5-minute interval, then X ~ Po(8.35).

(cor. to 4 sig.

fig.)

(b) Let Y be the random variable representing the number of cars passing through the tunnel in a t-minute interval and

is the mean of the distribution, then

Y ~ .

∴ The minimum vale of t is 0.8301. (cor. to 4 sig. fig.)

16. Let be the mean of the distribution.


(b)

(c)

17. Let X be the random variable representing the number of calls received per minute, then X ~ Po(3).(a)

(b)

Put n = 3, we have


Put n = 4, we have ∴ The minimum value of n is 4.

18. As there are on average 1.5 errors in typing 100 words, then there are on average 3.75 errors in typing 250 words.Let X be the random variable representing the number of errors in a particular typed page, then X ~ Po(3.75).(a)

(b) Let Y be the random variable representing the number of pages that need to be re-typed, then Y ~ B(20, 0.5162).

19. Let X be the random variable representing the number of students in Ken’s class receive grade A each year, then X ~ Po(4). Let Y be the random variable representing the number of students in Roger’s class receive grade A each year, then Y ~ Po(2).


(b)

(c)

(d)

20. (a) As there are on average 1 malfunction out of 50 machines each year, then there are on average 4 malfunctions out of 200 machines each year.Let X be the random variable representing the number of machines malfunction each year, then X ~ Po(4).


(b) As there are on average 1 malfunction out of 50 machines each year, then there are on average 0.4 malfunctions out of 20 machines each year.Let Y be the random variable representing the number of machines malfunction each year, then Y ~ Po(0.4).

(cor. to 4

sig. fig.)

As there are on average 1 malfunction out of 50 machines

each year, then there are on average malfunctions out of N

machines each year.Let Z be the random variable representing the number of

machines malfunction each year, then Z ~ Po .

(c)

(d)

21. Let X be the random variable representing the number of incoming calls per minute, then X ~ Po(2.5).

(a)

(b)

74


(c) As there are on average 2.5 incoming calls in a minute, then there are on average 5 incoming calls in 2 minutes.Let Y be the random variable representing the number of incoming calls in 2 minutes, then Y ~ Po(5).

22. Let X be the random variable representing the number of breakdowns in a year, then X ~ Po(4).(a)

(b) Expect maintenance fee

Exercise 12D (p. 12.45)Level 11. Let X be the random variable representing the number of

defective radios in each box, then X ~ B(5, 0.2).

(a)

(b)

(c) Let Y be the random variable representing the number of boxes among 10 boxes contain no defective radios, then Y ~ B(10, 0.32768).

2. (a) Let X be the random variable representing the number of under-filled bottles in a sample of 15 bottles, then X ~ B(15, 0.15).

(b)

Let Y be the random variable representing the number of samples out of 100 samples which contain fewer than 2 under-filled bottles, then Y ~ B(100, 0.979247783).

3. ∵ The team has the 4th win in their 8th match.∴ They win 3 times in the first 7 matches.The required probability

4. (a) As there are on average 1800 bacteria in 1 L of tap water, then there are on average 1.8 bacteria in 1 mL of tap water.Let X be the random variable representing the number of bacteria in 1 mL of tap water, then X ~ Po(1.8).

(b) Let Y be the random variable representing the number of contaminated specimens out of 5, then Y ~ B(5, 0.537163113).

5. (a) Let X be the random variable representing the number of cars entering a car park in a minute, then X ~ Po(1).


(b) Let Y be the random variable representing the number of 1-minute periods in the next 5 minutes in each of which more than 5 cars entering the car park, then Y ~ B(5, 0.000594184).

6. Let X be the random variable representing the number of substandard teddy bears in a bag of 5, then X ~ B(5, 0.03)

(a) (i) (cor. to 4 sig.

fig.)

(ii)

(b)∴ The total expected number of substandard teddy bears

in 100 bags is 100E(X) = 15.

Let Y be the random variable representing the number of the bags in a box containing no substandard teddy bears, thenY ~ B(10, 0.858734025).(c) (i)

(ii) (cor. to 4 sig. fig.)

(d) The total expected number of substandard teddy bears in 100 boxes is 1000E(X) = 150.

7. (a) As there are on average 1.5 flaws in a piece of cloth, then there are on average

flaws in a piece of cloth.


flaws in a piece of cloth, then X ~ Po .

(b) Let Y be the random variable representing the number of pieces of cloth out of 6 with more than 3 flaws, thenY ~ B(6, 0.004858176).

Level 28. (a) Let X be the random variable representing the number of

products tested until the first defective item is found, then

X ~ Geo (0.05).

(b) ∵ The 10th tested item is the 3rd defective item found.∴ There are total 2 defective items found in the first 9

tests.The required Probability

9. (a) Let X be the random variable representing the number of printing errors on a page, then X ~ Po(2).


(b) Let Y be the random variable representing the number of pages found until finding the page with no printing error,

then Y ~ Geo ( ).

(cor. to

4 sig. fig.)(c) Let Z be the random variable representing the number of

pages found out of 5 with no printing errors, then

Z ~ B(5, ).

(cor.

to 4 sig. fig.)

10. (a) Let X be the random variable representing the number of vehicles passing through the tunnel in a minute during peak hours, then X ~ Po(8).

(i) (cor. to 4

sig. fig.)

(ii)

(iii)

(b) Let Y be the random variable representing the number of 1-minute period out of 5 with not more than 4 vehicles passing through the tunnel, then Y ~ B(5, 0.0996324).

11. (a) Let X be the random variable representing the number of incoming calls between 10 am and 11 am, then X ~ Po(15).

76


(i) (cor. to

4 sig. fig.)

(ii)

(iii)

(b) Let Y be the random variable representing the number of working days out of 6 receiving exactly 15 incoming calls, then Y ~ B(6, 0.102435866).

12. Let X be the random variable representing the number of days required until the first break down occurs, then X ~ Geo (0.12).

(a) (cor. to

4 sig. fig.)

(b)

(c) ∵ The second breakdown occurs this Sunday.∴ There is a breakdown occurs before this Sunday.The required Probability

(d) ∵ The third breakdown occurs before this Sunday.∴ The third breakdown may occur on Wednesday,

Thursday, Friday, or Saturday.The required probability


goals in a match, then X ~ Po(2.5).

(a)

(b)

Let Y be the random variable representing the number of matches played until the first goalless match appears,

then Y ~ Geo ( ).

(c) ∵ The 5th match is the second goalless match.∴ There is one goalless match in the previous matches.The required probability

(d) Let Z be the random variable representing the number of

goalless matches in 10 matches, then Z ~ B(10, ).

14. Let X be the random variable representing the number of the weekly demand for cars, then X ~ Po(6.4).

(a) (i) (cor. to

4 sig. fig.)

(ii)

(iii)

(b) Consider the probability function:


For ,

when n = 9, ;

when n = 10, .∴ The minimum number of cars the company should

keep is 10.

(c) Let Y be the random variable representing the number of weeks out of 5 in each of which there is a demand for more than 4 cars, then Y ~ B(5, 0.764929966).


Revision Exercise 12 (p. 12.50)Level 11. (a) (i) y 0 1

P(Y = y)

(ii) , where y = 0, 1.

(b)

2. (a) R = 0, 1

(b) , where r = 0, 1.

(c)

3. Let X be the random variable representing the number of bulbs out of 20 working properly for more than 1 year, then X ~ B(20, 0.15).(a)

(b) The required number

4. Let X be the random variable representing the number of confirmed reservations out of 7, then X ~ B(7, 0.4).(a)

(b) (cor. to 4 sig. fig.)

(c) E(X) = 7 0.4 = 2.8E (number of passengers)

5. Let X and Y be the random variables representing the number of questions which Jack and Wendy guess correctly respectively, then X ~ B(6, 0.2) and Y ~ B(4, 0.2).

6. Let X be the random variable representing the number of sunny days in April, then X ~ B(30, 0.7).(a)

(b)

(c)

(d) P(rainy day | shopping)

7. (a) Let X be the random variable representing the number of heads occur out of 5, then X ~ B(5, 0.7).

(b) Let Y be the random variable representing the number of tosses required until the first head occurs, then Y ~ Geo(0.7).

78


8. Let X and Y be the random variables representing the number of times required until the first successful jump occurs on a dry day and a wet day respectively, then X ~ Geo(0.75) and Y ~ Geo(0.5).

(a) (cor. to

4 sig. fig.)

(b)

(c) (i)

(ii) E(number of jumps on a certain day)

9. Let X be the random variable representing the number of years between 2 successive earthquake, then X ~ Geo(0.005).

10. (a) (i) Let X be the random variable representing the number of cards drawn until the first diamond card is drawn, then X ~ Geo(0.25).

(ii) Let Y be the random variable representing the number of diamonds drawn out of 10, then Y ~ B(10, 0.25).

(b)

(c)

11. (a) Let X be the random variable representing the number of cars entering the gas station in a minute, then X ~ Po(3).

(b) Let Y be the random variable representing the number of minutes out of 5 in each of which more than 5 cars enter the gas station, then Y ~ B(5, 0.083917942).

(c) Let Z be the random variable representing the number of minutes required until there is at least a car cannot be served in a minute, then Z ~ Geo(0.083917942).

12. For a Poisson distribution, the variance is equal to the mean. Thus there are on average 2 defective components daily, and there are on average 6 defective components during a 3-day period.Let X be the random variable representing the number of defective components during a three-day period, then X ~ Po(6).

13. (a) As there is an average rate of 2.5 accidents every week, there is an average rate of 5 accidents every fortnight.Let X be the random variable representing the number of accidents per fortnight, then X ~ Po(5).

(b) (cor. to 4 sig. fig.)

14. (a) Let X be the random variable representing the number of claims on a day, then X ~ Po(4.5).

(b) Let Y be the random variable representing the number of days out of 5 in each of which it will handle more than 2 claims, then Y ~ B(5, 0.826421929).

15. (a) As there is on average 1 typing error per 2 pages, then there is on average 0.5 typing error per page.Let X be the random variable representing the number of typing errors in a page, then X ~ Po(0.5).


(b) Let Y be the random variable representing the number of pages required until the first page free of typing errors is

found, then Y ~ Geo( ).

16. Let X be the random variables representing the number of colds Calvin gets in a year, then X ~ Po(2.5).If the numbers of colds that year of Calvin are more than or equal to 1, it has a probability of 0.8 to reduce to 1 after taking that medicine.The required probability

Level 217. Let X be the random variable representing the number of

rotten apples found in the sample, then X ~ B(20, 0.05).P(the lot of apples is rejected)

P(the sample contains 3 rotten apples)

P(the sample contains 3 rotten apples | the lot of apples is rejected)

18. (a) Let n be the total number of the batteries and p be the probability of producing a defective batteries. Then X ~ B(n, p).E(X) = np = 4……(1)Var (X) = np (1 – p) = 3.84……(2)By solving (1) and (2), we have1 – p = 0.96

p = 0.04 and n = 100

∴ , where x = 0,

1, …, 100(b)

19. Let n be the number of missiles to be launched.Let X be the random variables representing the number of missiles hit the target, then X ~ B(n, 0.4).

∴ The minimum number of missiles to be launched is 5.

20. (a)

(b)

21. (a) Let X be the random variables representing the number of girls in a 4-child family, then X ~ B(4, 0.6).

(i)

(ii)

(iii)

(b)

(c) Let Y be the random variables representing the number of boys in a family, then Y ~ B(n, 0.4), where n is the total number of children in the family.

∴ The minimum number of children is 6.

22. Let X and Y be the random variables representing the number of defective watches drawn in Step I and Step II respectively, then X ~ B(10, 0.1) and Y ~ B(20, 0.1).(a) P(2nd sample is drawn and the whole batch is rejected)

(b) P(accept whole batch)

80


(c)

23. Let X be the random variables representing the number of jurors make correct decision, then X ~ B(n, p), where n is the total number of jurors.(a) (i) When n = 1, we have

(ii) When n = 3, we have

(b) If p > 0.5, the probability of making a correct final decision will be higher with more jury members, and vice versa (or any reasonable answer).

24. Let X and Y be the random variables representing the number of heads obtained if coin A and coin B are selected respectively, then X ~ B(4, 0.5) and Y ~ B(4, 0.65),

(a)

(b) P(coin A | 1 head in 4 tosses)

(c) P(head is obtained in first toss | coin A is tossed and 1 head in 4 tosses)

(d) P(coin A is tossed and head is obtained in first toss | 1

head in 4 tosses)

25. Let X and Y be the random variables representing the number of cars and motorcycles served by the petrol station in an hour respectively, then X ~ Po(3.5) and Y ~ Po(1.5).

26. Let X be the random variables representing the number of experiments performed, then X ~ Geo(0.25).

(a)

E(cost of the project)

(b)

27. (a) Let X be the random variable representing the number of flaws in 1 cm of copper wire, then X ~ Po(2.3).


(b) As there are on average 2.3 flaws in 1 cm of copper wire, then there are on average 4.6 flaws in 2 cm of copper wire.Let Y be the random variable representing the number of flaws in 2 cm of copper wire, then Y ~ Po(4.6).


(c) As there are on average 2.3 flaws in 1 cm of copper wire, then there are on average 1.15 flaws in 0.5 cm of copper wire.Let Z be the random variable representing the number of flaws in 0.5 cm of copper wire, then Z ~ Po(1.15).

(d)

Let N be the random variable representing the number of parts out of 10 which have no flaws, then

N ~ B(10, ).

28. Let X be the random variable representing the number of cars demand for renting per week, then X ~ Po(5.4).(a) If the company keeps 5 cars, we have

E(weekly profit)

If the company keeps 6 cars, we haveE(weekly profit)

(b) ∵ The expected weekly profit of keeping six cars is greater than keeping 5 cars.

∴ The company should keep 6 cars in order to maximize the weekly profit.

29. (a) Let X be the random variable representing the number of particles emitted in the next second, then X ~ Po(2.8).


(b) As there are on average 2.8 particles emitted in a second, then there are on average 8.4 particles emitted in 3 seconds.Let Y be the random variable representing the number of particles emitted in 3 seconds, then Y ~ Po(8.4).

(c) As there are on average 2.8 particles emitted in a second, then there are on average 2.8n particles emitted in n seconds.Let Z be the random variable representing the number of particles emitted in n seconds, then Z ~ Po(2.8n)

When n = 2, 0.80937

When n = 3, 0.96774∴ The minimum value of n is 3.

30. (a) Let X be the random variable representing the number of trucks among 10 which are overloaded, then X ~ B(10, 0.05).

82


(b) Let Y be the random variables representing the number of trucks inspected until the first overloaded truck appears, then Y~ Geo(0.05).

(cor. to 4 sig.

fig.)

(c) ∵ The 10th truck is the third overloaded truck.∴ There are 2 overloaded trucks in the previous 9 trucks.The required probability

31. (a) It is given that X ~ Po(1.6).

Let Y be the random variable representing the number of

samples out of 10 which contain no fish, then Y ~ B(10,

(b)

Let Z be the random variable representing the number of samples out of 10 which contain more than 1 fish, then Z ~ B(10, 0.475069053).

(c)


32. Let X be the random variable representing the number of errors on a page, then X ~ Po(2).


(b) E(number of errors) =

(c)

Let Y be the random variable representing the number of pages out of 300 contain more than 3 errors, then Y ~ B(300, 0.142876539).

(d) E(number of pages containing at least 1 error)


Challenging Questions (p. 12.56)

1. (a) As there are on average 2.5 accidents per month, then

there are on average accidents per year.

(b) As there are on average 2.5 accidents per month, then there are on average accidents per 2 months.Let X be the random variable representing the number of accidents in 2 months, then X ~ Po(5).


(c) Let Y be the random variable representing the number of months elapsed until no accidents appear in a month, then

Y ~ Geo( ).

(cor. to 4

sig. fig.)

(d) Let Z be the random variable representing the number of months out of 12 in each of which no accidents appear,

then Z ~ B(12, ).

2. (a) Let X and Y be the random variables representing thenumber of cars using tunnels A and B per minute respectively, then X ~ Po(3) and Y ~ Po(2),

(b)

(c) As there are on average 3 and 2 cars using tunnel A and B per minute respectively, then there are on average 6 and 4 cars using tunnel A and B per 2 minutes respectively.Let M and N be the random variables representing thenumber of cars using tunnels A and B per 2 minutes respectively, then M ~ Po(6) and N ~ Po(4),


(d) and

Let C be the random variable representing the number of 1 minute period out of 5 in each of which no cars use

tunnel A, then C ~ (5, ).

Let D be the random variable representing the number of 1 minute period out of 5 in each of which no cars use

tunnel B, then D ~ (5, ).

(e)

Let Z be the random variable representing the number of 1 minute period out of 5 in each of which no cars use

tunnel A and B, then Z ~ B(5, ).

84

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