11类RD Sharma解决方案–第33章概率–练习33.3 |套装1

问题1.以下哪一项不能有效分配样本空间结果基本事件的概率S = {w ₁ ，w ₂ ，w ₃ ，w ₄ ，w ₅ ，w ₆ ，w ₇ }：

基本事件是：

	w1	w2	w3	w4	w5	w6	w7
1	0.1	0.01	0.05	0.03	0.01	0.2	0.6
2	1/7	1/7	1/7	1/7	1/7	1/7	1/7
3	0.7	0.6	0.5	0.4	0.3	0.2	0.1
4	1/14	2/14	3/14	4/14	5/14	6/14	15/14

解决方案：

We know that  the value of probability of an event must be between 0 and 1 and that the sum of all probabilities of a event must be equal to one. Using this rationale,

1. This is valid since each probability of event w₁ lies between 0 and 1 and the sum of all probabilities of event w₁ is 1, or P{w₁}= 1.

2. This is valid since each probability of event w₂ lies between 0 and 1 and the sum of all probabilities of event w₂  is 1, or P{w₂}=2.

3. This is not valid as the sum of all probabilities is 2.8, which is more than one. P{w_i} = 2.8

4. This is not valid since the probability of w₇ is 15/14= 1.07 > 1.

Hence, the valid events are only 1 and 2.

问题2.掷骰子。找到获得的可能性：

(i)质数

解决方案：

Since a die is thrown, the number of outcomes in sample space must be 6. Thus, n{S}= 6.

Let E be the event of getting a prime number.

∴ E = {2,3,5} or n{E} = 3

We know, Probability of an event = Number of outcomes in that event / number of outcomes in sample space

∴ P{E} = n{E}/ n{S} = 3/6

∴ P{E}= 1/2

(ii)2或4

解决方案：

Let N be the event of getting a 2 or 4.

Hence, N = (2,4} or n{N} = 2

Probability of event N or P{N} = n{N}/n{S}

or P{N} = 2/6

∴ P{N} = 1/3

(iii)2或3的倍数

解决方案：

Let R be the event of getting a multiple of 2 or 3 when a die is thrown.

∴ R = {2,3,4,6}

Thus, n{R} = 4

∴ P{R} = n{R}/ n{S} = 4/6

∴ P{R} = 2/3

问题3：在掷骰子的同时掷骰子中，求出以下几率：

(i)总和为8

解决方案：

Since a pair of die have been thrown, the total number of outcomes in the sample space S become

6×6 = 6²= 36.

Let R be the event of getting 8 as the sum when a pair of dice is thrown together.

∴ R = {(2,6), (3,5), (4,9), (5,3), (6,2)}

∴ n{R} = 5

Hence probability of R = n{R}/ n{S}

∴ P{R} = 5/36

(ii)一双

解决方案：

Let D be the event that a doublet appear on the dice when a pair of dice is thrown together.

∴ D = { (1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}

Thus, n{D} = 6

Hence probability of D = n{D}/ n{S}

P{D} = 6/36 = 1/6

∴ P{D} = 1/6

(iii)质数的对偶

解决方案：

Let F be the event of getting a doublet of prime numbers when a pair of dice is thrown together.

∴ F = {(2,2), (3,3), (5,5)}

Thus, n{F} = 3

Hence probability of F = n{F}/ n{S}

P{F}= 3/36 = 1/12

∴ P{F} = 1/12

(iv)偶数的双数

解决方案：

Let Q be the event of getting a doublet of odd numbers when a pair of dice is thrown.

∴ Q = {(1,1), (3,3), (5,5)}

Thus, n{Q} = 3

 Probability of Q = n{Q}/ n{S}

P{Q} = 3/36 = 1/12

∴ P{Q} = 1/12

(v)总和大于9

解决方案：

Let K be the event of getting a sum greater than 9 when a pair of dice is thrown.

∴ K = {(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)}

Thus, n{K} = 6

Probability of K = n{K}/n{S}

P{K}= 6/36 = 1/6

∴ P{K} = 1/6

(vi)首先是偶数

解决方案：

Let W be the event of getting an even number on first dice when a pair of dice is thrown. If the first number has to be even, it means that any other number can appear on the second dice, be it even or odd. Using this rationale:

∴ W ={(2,1), (2,2), (2,3), (2,4),(2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

 Thus, n{W} = 18

 Probability of W = n{W}/ n{S}

= 18/36 = 1/2

∴ P{W} = 1/2

(vii)一个是偶数，另一个是3的倍数

解决方案：

Let V be the event of that we get an even number on one and a multiple of 3 on the other die when a pair of dice is thrown simultaneously. In this case, it is not specified that the even number shall have to be on the face of first or the second dice.

So long as an even number appears on any of the two dices, it is to be taken as an outcome of the given event. Same goes for multiple of 3 as well. Using this rationale:

∴ V = {(2,3), (2,6), (4,3), (4,6), (6,3), (6,6), (3,2), (3,4), (3,6), (6,2), (6,4)}

Thus, n{V} = 11

Probability of V = n{V}/ n{S}

∴ P{V} = 11/36

(viii)9或11都不是面孔上数字的总和

解决方案：

Let H be the event of getting neither 9 nor 11 as the sum of the numbers on the faces of two dice when they are thrown together.

Therefore, H’ shall be the event that either 9 or 11 appear as sum of the two numbers on the faces of the two dices.

H’ = {(3,6), (4,5), (5,4), (5,6), (6,3), (6,5)}

n{H’} = 6

P{H’} = n{H’}/ n{S} = 6/36 = 1/6                  

Thus, P{H} = 1−P{H’} = 1−1/6 = 5/6                            

(ix)少于6的款项

解决方案：

Let E be the event of obtaining a sum less then 6 when a pair of die is thrown together.

∴ E = {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (4,1)}

∴ n{E} =10

Probability of E = P{E} = n{E}/ n{S} = 10/36 = 5/18

∴ P(E) = 5/18

(x)少于7的总和

解决方案：

Let C be the event that a sum less than 7 appears on the faces of the two dice when thrown together.

∴ C = {(1,1), (1,2), (1,3), (1,4), (1,5), (2,1), (2,2),(2,3),(2,4),(2,5),(3,1),(3,2),(3,3).(4,1),(4,2),(5,1)}

Thus, n{C} = 15

Probability of C = P{C} = n{C}/ n{S}

= 15/36 = 5/12

∴ P(C) = 5/12

(xi)一笔多于7的款项

解决方案：

Let X be an event of getting a sum more than 7 from the numbers on the faces of the two dice when thrown together.

∴ X = {(2,6), (3,5), (3,6), (4,4), (4,5), (4,6), (5,3),(5,4), (5,5), (5,6), (6,2), (6,3), (6,4), (6,5), (6,6)}

Thus n{X} = 15

Probability of X = P{X} = n{X}/ n{S} = 15/36 = 5/12

∴ P(X) = 5/12

(xii)既不是双胞胎也不是10

解决方案：

Let A be the event of getting neither a doublet nor a total of 10 when a pair of dice is thrown simultaneously.

∴ A’ is the event of getting either a doublet or a total of 10 on the faces of the two dices.

∴ A’ = {(1,1), (2,2), (3,3), (4,6), (5,5), (6,4), (6,6)}

Thus, n{A’} = 8

Probability of A’ = n{A’}/ n{S} = 8/36 = 2/9

Probability of A = P{A} = 1−P{A’}

= 1−2/9

∴ P(A’} = 7/9

(xiii)第一个为奇数，第二个为6

解决方案：

Let B be the event of getting an odd number on the first and 6 on the second face of the two dices.

∴ B = {(1,6), (3,6), (5,6)}

Thus n{B} = 3

Probability of B = P{B} = n{B}/ n{S} = 3/36 = 1/12

∴ P(B} = 1/12

(xiv)每边大于4的数字

解决方案：

Let J be the event of getting a number greater than 4 on each side of dice when both dices are thrown simultaneously.

∴ J = {(5,5), (5,6), (6,5), (6,6)}

Thus n{J} = 4

Probability of J = P{J} = n{J}/ n{S} = 4/36 = 1/9

∴ P(J} = 1/9

(xv)总计9或11

解决方案：

Let I be the event of getting a total of 9 or 11 when two dice are thrown.

∴ I = {(3,6),(4,5),(5,4),(5,6),(6,3),(6,5)}

Thus n{I} = 6

Probability of I = P{I} = n{I}/ n{S} = 6/36 = 1/6

∴ P(I} = 1/6                                  

(xvi)总数大于8

解决方案：

Let Z be the event of getting a total greater than 8.

∴ Z = {(3,6),(4,5),(4,6),(5,4),(5,5),(5,6),(6,3),(6,4),(6,5),(6,6)}

∴ n{Z} = 10

Probability of Z = n{Z}/ n{S} = 10/36 = 5/18              

∴ P(Z} = 5/18                                                  

问题4.一掷三骰子，求出总共得到17或18的概率。

解决方案：

Since three dice have been thrown simultaneously, the total number of outcomes in that sample space become 6×6×6= 6³ = 216.

Let E be the event of getting a total of 17 or 18. Thus E = {(6,6,5), (6,5,6), (5,6,6), (6,6,6)}

Hence, n{E} = 4

Probability of event E = P{E} = n{E}/ n{S} = 4/216 = 1/54

∴ P{E} = 1/54

问题5.将三枚硬币扔在一起。找到获得的可能性：

(i)正好2个头

解决方案：

Since 2 coins have been tossed together, the total outcomes of the sample space are n{S} =

 2×2×2 = 2³ = 8.

Let A be the event of getting exactly 2 heads.

∴ A = {HHT, HTH, THH} or, n{A} = 3            

Probability of event A = P{A} = n{A}/n{S} = 3/8    

∴ P{A} = 3/8

(ii)至少两个首长

解决方案：

Let B be the event of getting at least 2 heads when three coins are thrown together.

∴ B = {HHH, HHT, THH, HTH} or, n{B} = 4

Probability of event B = P{B} = n{B}/ n{S} = 4/8 = 1/2

∴ P{B} = 1/2

(iii)至少一头一尾

解决方案：

Let C be the event of getting at least one head and one tail .

C= {(HHT, THT, HTT, TTH, HTH, THH} or, n{C}= 6

P{C}= n{C}/ N(S) = 6/8 = 3/4

∴ P{C}= 3/4      

问题6.普通年份有53个星期日的概率是多少?

解决方案：

We know that an ordinary year has 52 weeks and 1 day.

52 weeks in a year obviously implies 52 Sundays.

But in the question, we have to find the probability of having 53 Sundays which means that we have to fine the probability of the last 1 day of an ordinary year to be a Sunday.

Total number of days in a week = Number of outcomes of sample space = n(S) = 7

S= {MONDAY, TUESDAY, WEDNESDAY, THURSDAY, FRIDAY, SATURDAY, SUNDAY}

Hence, the probability of that one day being a Sunday = 1/7

Thus, the probability of having 53 Sundays in an ordinary year is 1/7.

问题7. year年有53个星期日和53个星期一的概率是多少?

解决方案：

We know that in a leap year we have 52 weeks and 2 days (366 days)

The sample space for the last 2 days shall be

S= {(Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday), (Sunday, Monday)} or, n{S}=7

Let E denote the event of getting a Sunday and a Monday as the last 2 days in a leap year.

∴ E= {Sunday, Monday} or, n{E} = 1

Probability of E= P{E}= n{E}/ n{S} = 1/7

Thus, the probability that a leap year has 53 Sundays and 53 Mondays is 1/7.

问题8.一个袋子包含8个红色和5个白色的球。随机抽出三个球。找到以下可能性：

(a)所有三个球都是白色的

解决方案：

Since there are 8+5= 13 balls in the bag and we are required to draw any three balls at random.

Total number of outcomes shall be= n{S}= ¹³C₃ = 286.

Let A be the event that all the three balls drawn are white.

Hence the number of outcomes in event A= n{A} = ⁵C₃= 10.

Probability of event A= n{A}/n{S} = 10/286= 5/143

Hence the probability of drawing all three white balls is 5/143.

(b)所有三个球都是红色的

解决方案：

Let R denote the event that all three balls drawn are red. 

Since total number of red balls is 8, the number of ways of drawing 3 red balls out of 8 = ⁸C₃ = 56 = n{R}.

We know, n{S}= 286

∴ Probability of R= n{R}/n{S}= 56/286 = 28/143

Hence, the probability of drawing all three red balls is 28/143.

(c)一个球为红色，两个球为白色

解决方案：

Let E be the event that one ball drawn is red and the other two balls are white.

Number of outcomes of E= ⁸C₁ × ⁵C₂ = 8 ×10 = 80.

∴ Probability of E= n{E}/n{S} = 80/286 = 40/143

Hence the probability of getting one red and other two balls as white is 40/143.

问题9.在三个骰子的单掷中，找出在所有三个骰子上获得相同数字的概率。

解决方案：

Since three dice have been rolled together, the total number of outcomes in the sample space= n{S}= 6³= 6×6×6= 216.

Let E be the event of getting the same numbers on all the three dice.

∴ E= {(1,1,1), (2,2,2), (3,3,3), (4,4,4), (5,5,5), (6,6,6)}

Number of outcomes in E= n{E} = 6

Probability of E= n{E}/n{S}= 6/216 = 1/36

Hence the probability of getting the same numbers on all the three dice is 1/36.

问题10：将两个不偏不倚的骰子扔到一起。找出骰子上数字总数大于10的概率。

解决方案：

Since two dice have been rolled together, the total number of outcomes in the sample space= n{S}= 62= 6×6= 36.

Let E be the event of getting a total of numbers on the dice greater than 10.

∴E= {(5,6), (6,5), (6,6)}

Number of outcomes in E= n{E}= 3

Since n{S}= 52, Probability of E= n{E}/n{S}= 3/36 = 1/12

Hence the probability of getting a total of numbers on the dice greater than 10  is 1/12.

问题11：从52张纸牌中随机抽取一张纸牌。找出开牌的概率为：

(i)黑人国王

解决方案：

Since  a card is drawn from a pack of 52 cards, so number of elementary events in the sample space = n {S} = ⁵²C₁ = 52.

Let E be the event of drawing a black king. Since there are two black kings, one of spade and other of club,

∴n {E} = ²C₁ = 2

Since n{S}= 52, Probability of event E = n{E}/n{S} = 2/52 = 1/26.

Hence the probability of drawing a black king is 1/26.

(ii)黑牌或国王

解决方案：

Let A be the event of drawing a black card or a king. We know that there are 26 black cards and 4 kings out of which 2 kings are black.

Hence, total number of outcomes of event A= n{E} = ²⁶C₁ + ⁴C₁ – ²C₁= 28

In this case, 2 needs to be subtracted from total because there are two black kings which are already counted in black cards and to avoid double counting.

Since n{S}= 52, Probability of A= n{A}/n{S} = 28/52 = 7/13.

Hence, the probability of getting either a black card or a king is 7/13.

(iii)黑与王

解决方案：

Let W be the event of drawing a black card and a king. We know there are two black kings, one of spade and other of club.

Hence number of outcomes of W= n{W} = ²C₁ = 2.

Since n{S}=52, Probability of W= n{W}/n{S} = 2/52 = 1/26.

Hence the probability of drawing a black card and a king is 1/26.

(iv)千斤顶，皇后或国王

解决方案：

Let B be the event of drawing a jack, queen or king. We know there are 4 kings, 4 queens and 4 jacks in a deck of cards.  

Hence total number of outcomes of event B = n{B} = ⁴C₁ + ⁴C₁ + ⁴C₁ = 12.

Since n{S}= 52, Probability of B = P{B} = n{B}/n{S} = 12/52 = 3/13

Hence the probability of drawing a jack, queen or king is 3/13.

(v)既不是一颗心也不是国王

解决方案：

Let L be the event of drawing neither a heart nor a king and let L′ as the event that either a heart or king appears.

Since there are 13 hearts and 4 kings total number of outcomes = n (L′) = ⁶C₁ + ⁴C₁ – 1=16 ,deducting the king of hearts.

Probability of L’ = P{L′} = n{L’}/n{S} = 16/52 = 4/13

So, P{L} = 1 – P{L’} = 1 – 4/13 = 9/13

Hence the probability of drawing  neither a heart nor a king is 9/13.

(vi)锹或王牌

解决方案：

Let D be the event of drawing a spade or king. We know there are 13 spades and 4 kings, but 1 king is already included in the 4 kings.

Number of outcomes in D = n{D} = ¹³C₁ + ⁴C₁ – 1=16

Since N{S} = 52, Probability of event D = P{D} = n{D}/n{S} = 16/52 = 4/13

Hence the probability of drawing either a spade or an ace is 4/13.

(vii)既不是王牌也不是国王

解决方案：

Let K be the event of drawing neither an ace nor a king and K′ as the event that either an ace or king appears.

Number of outcomes in K’ = n{K’} = ⁴C₁ + ⁴C₁ = 8

Since n{S}= 52, Probability of K’ = P{K’} = n{K’}/n{S} = 8/52= 2/13

So, P{K} = 1 – P{K’} = 1 – 2/13= 11/13

Hence the probability of drawing neither an ace nor a king is 11/13.

(viii)钻石卡

解决方案：

Let X be the event of drawing a diamond card. We know there are 13 diamond cards.

Hence number of outcomes of event X = n{X} = ¹³C₁ = 13.

Since n{S} = 52, Probability of event X = P{X} = n{X}/n{S} = 13/52 = 1/4

Hence the probability of drawing a diamond card is 1/4.

(ix)不是钻石卡

解决方案：

Let X be the event of drawing not a diamond card and X′ as the event that diamond card appears. We know there are 13 diamond cards.

Hence number of outcomes of event X = n{X} = ¹³C₁ = 13.

Since n{S} = 52, Probability of event X = P{X} = n{X}/n{S} = 13/52 = 1/4

So, P{X’} = 1 – P{X’} = 1 – 1/4 = 3/4

Hence the probability of not drawing a diamond card is 3/4.

(x)黑卡

解决方案：

Let E be the event of drawing a black card. We know there are 26 black cards (spades and clubs).

Hence total number of outcomes of event E = n{E} = ²⁶C₁ = 26

Since n{S} = 52, Probability of event E = P{E} = n{E}/n{S} = 26/52= 1/2

Hence the probability of drawing a black card is 1/2.

(xi)不是王牌

解决方案：

Let E be the event of drawing not an ace and  E′ as the event that ace card appears. We know that there are 4 ace cards.

Number of outcomes in E’ = n{E’} = ⁴C₁ = 4.

Since n{S} = 52, Probability of event E’ = P{E’} = n{E’}/n{S} = 4/52 = 1/13

So, P (E) = 1 – P (E′) = 1 – 1/13 =12/13

Hence the probability of not drawing an ace is 12/13.

(xii)不是黑卡

解决方案：

Let E be the event of not drawing a black card. We know there are 26 cards other than black cards (red cards of hearts and diamonds)

Hence number of outcomes of drawing a red card = n{E} = ²⁶C₁ = 26 .

Since n{S} = 52, Probability of event E = P{E} = n{E}/n{S} = 26/52 = 1/2.

Hence the probability of not drawing a black card is 1/2.

问题12.在洗刷一包52张纸牌时，有四张被意外掉下。找到每套西装中缺牌的几率。

解决方案：

A pack of 52 cards from which 4 are dropped. We now have to find the probability that the missing cards should be one from each suit.

We know that, from well shuffled pack of cards, 4 cards missed out total possible outcomes = n{S} = ⁵²C₄ = 270725

Let E be the event that four missing cards are from each suite

Hence total number of outcomes of event E = n{E} = ¹³C₁ × ¹³C₁ × ¹³C₁ × ¹³C₁ = 134

Probability of event E =P{E} = n{E}/n{S} = 134/270725 = 2197/20825

Hence the probability that the missing cards should be one from each suit is 2197/20825.

问题13.从一副52张纸牌中，同时抽出四张纸牌。找到他们将成为同一套西装的四项荣誉的机会。

解决方案：

We have to find the probability that all the face cards  drawn from a pack of 52 cards are of same suits.

Total possible outcomes in sample space =n{S} = ⁵²C₄

Let E be the event that all the cards drawn are face cards of same suit.

Hence number of outcomes of event E = n (E)= 4 × ⁴C₄ = 4 × 1 = 4.

Probability of event E =P{E} = n{E}/n{S}= 4/270725

Hence the probability that the face cards  drawn from a pack of 52 cards are of same suits is 4/270725.

问题14.从1到20的票证混合在一起，然后随机抽取一张票证。票证的数字是3或7的倍数的概率是多少?

解决方案：

We have to find the probability that the ticket has a number which is a multiple of 3 or 7.

Total possible outcomes in sample space =n{S} = ²⁰C₁ = 20.

Let E be the event of getting ticket which has number that is multiple of 3 or 7.

Hence outcomes of event E are {3,6,9,12,15,18,7,14}. Number of outcomes of event E are n{E} = 8.

Probability of event E =P{E} = n{E}/n{S}= 8/20 = 4/5

Therefore, the probability that the ticket has a number which is a multiple of 3 or 7 is 4/5.

问题15.一个袋子包含6个红色，4个白色和8个蓝色的球。如果随机抽出三个球，则求出一个概率是红色，一个是白色，一个是蓝色的概率。

解决方案：

We have to find the probability that the one is red, one is white and one is blue.

Since three balls are drawn so, total number of outcomes for drawing 3 balls is n{S} = ¹⁸C₃ = 816

Let E be the event that one red, one white and one blue ball is drawn.

n{E} = ⁶C₁ × ⁴C₁ × ⁸C₁ = 192

Probability of event E =P{E} = n{E}/n{S} = 192 / 816 = 4/17

Therefore, the probability that one is red, one is white and one is blue is 4/17.