11类RD Sharma解决方案–第33章概率–练习33.3 |套装3

问题31.一个袋子包含5个红色，6个白色和7个黑色的球。随机抽出两个球。找出两个球都为红色或黑色的概率。

解决方案：

Clearly the bag has 5 + 6 + 7 = 18 balls and since 2 balls have been drawn at random, number of outcomes in sample space = n{S} = ¹⁸C₂ = 153

Let E be the event that the two balls chosen are either red or black.

Hence number of outcomes in event E = n{E} = ⁵C₂ + ⁷C₂ = 31

Probability of E = n{E}/n{S} = 31/153

Hence the probability of both balls being red or black is 31/153.

问题32.如果从英语字母表中随机选择一个字母，则求出该字母是：

(i)元音

解决方案：

We know there are 26 letters in English alphabet. Since one letter has been chosen from the 26 letters, number of outcomes in sample space = n{S} = ²⁶C₁ = 26

Let V be the event that the chosen letter is a vowel. Obviously there are 5 vowels in the English alphabet.

Hence the number of outcomes favorable to event V = n{V} = ⁵C₁ = 5.

Probability of event V = n{V}/n{S} = 5/26.

Hence the probability of getting a vowel is 5/26.

(ii)辅音

解决方案：

Let C be the event that the chosen letter is a consonant. Obviously there are 21 consonants in the English alphabet.

Hence the number of outcomes favorable to event C = n{C} = ²¹C₁ = 21.

Since n{S} = 26, Probability of event C = n{C}/n{S} = 21/26.

Hence the probability of getting a consonant is 21/26.

问题33.在彩票中，一个人从1到20中选择6个不同的数字，如果这6个数字与彩票委员会确定的6个数字匹配，那么他将中奖。赢得比赛的几率是多少?

解决方案：

Since there are 20 numbers, and 6 numbers have to be chosen, the number of outcomes in sample space = n{S} = ²⁰C₆ = 38760

Let L be the event that the 6 numbers chosen match with the already decided numbers. We know that there is only 1 set pattern to the numbers that have already been fixed by the lottery committee.

Hence number of outcomes favorable to event L = n{L} = 1.

Probability of event L = n{L}/n{S} = 1/38760.

Hence the probability of winning the lottery is 1/38760.

问题34. 20张卡的编号从1到20。随机抽取一张卡。找出卡片上的数字是：

(i)4的倍数

解决方案：

Since 1 card has been drawn from 20 given cards, number of outcomes in sample space = n{S}

=> ²⁰C₁ = 20.

Let Y be the event that the number on the card is a multiple of 4. We know there are 5 multiples of 4 from 1 to 20, (i.e., 4, 8, 12, 16, 20.)

Hence number of outcomes favorable to Y = n{Y} = 5

Probability of event Y = P{Y} = n{Y}/n{S} = 5/20 = 1/4.

Hence the probability of the number on the card being a multiple of 4 is 1/4.

(ii)不是4的倍数

解决方案：

Let Y be the event that the number on the card is a multiple of 4 and Y’ denote the event that it is not a multiple of 4.

We know there are 5 multiples of 4 from 1 to 20, (i.e., 4, 8, 12, 16, 20.)

Hence number of outcomes favorable to Y = n{Y} = 5

Since n{S} = 20, Probability of event Y = n{Y}/n{S} = 5/20 = 1/4.

Thus, P{Y’} = 1 − P{Y} = 1 − 1/4 = 3/4.

Hence the probability of the number on the card not being a multiple of 4 is 3/4.

(iii)奇数

解决方案：

Let O be the event that the number on the card is odd. We know there are 10 odd numbers from 1 to 20, (i.e., 1, 3, 5, 7, 9, 11, 13, 15, 17, 19.)

Hence number of outcomes favorable to O = n{O} = 10

Since n{S} = 20, Probability of event O = P{O} = n{O}/n{S} = 10/20 = 1/2.

Hence the probability of the number on the card being an odd number is 1/2.

(iv)大于12

解决方案：

Let G be the event that the number on the card is odd. We know there are 8 numbers greater than 12 from 1 to 20, (i.e., 13, 14, 15, 16, 17, 18, 19, 20.)

Hence number of outcomes favorable to G = n{G} = 8

Since n{S} = 20, Probability of event G = P{G} = n{G}/n{S} = 8/20 = 2/5.

Hence the probability of the number on the card being greater than 12 is 2/5.

(v)可除以5

解决方案：

Let F be the event that the number on the card is divisible by 5. We know there are 4 numbers divisible by 5 from 1 to 20, (i.e., 5, 10, 15, 20.)

Hence number of outcomes favorable to F = n{F} = 4

Since n{S} = 20, Probability of event F = P{F} = n{F}/n{S} = 4/20 = 1/5.

Hence the probability of the number on the card being divisible by 5 is 1/5.

(vi)不是6的倍数

解决方案：

Let Z be the event that the number on the card is a multiple of 6 and Z’ denote the event that it is not a multiple of 6.

We know there are 3 multiples of 6 from 1 to 20, (i.e., 6, 12, 18.)

Hence number of outcomes favorable to Z = n{Z} = 3

Since n{S} = 20, Probability of event Z = n{Z}/n{S} = 3/20

Thus, P{Z’} = 1 − P{Z} = 1 − 3/20 = 17/20.

Hence the probability of the number on the card not being a multiple of 6 is 17/20.

问题35.投掷两个骰子。找出赔率：

(i)赞成获得总和4

解决方案：

Since a pair of dice has been thrown, number of items in sample space = n{S}= 62 = 36.

Let E be the event of getting a sum of 4 on the dice and E’ be the event of not getting a sum of 4.

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

Hence number of outcomes favorable to event E = n{E} = 3

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

Now, P{E’} = 1 − P{E} = 1 − 1/12 = 11/12.

Thus odds in favour of getting a sum of 4 = P{E}/P{E’} = 1:11.

(ii)赞成求和5

解决方案：

Let C be the event of getting a sum of 5 on the dice and E’ be the event of not getting a sum of 5.

C = {(1,4), (2,3), (3,2), (4,1)}

Hence number of outcomes favorable to event C = n{C} = 4

Since n{S} = 36, Probability of event C = P{C} = n{C}/n{S} = 4/36 = 1/9

Now, P{C’} = 1 − P{C} = 1 − 1/9 = 8/9.

Thus odds in favour of getting a sum of 5 = P{C}/P{C’} = 1:8.

(iii)反对获得6的总和?

解决方案：

Let A be the event of getting a sum of 6 on the dice and A’ be the event of not getting a sum of 6.

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

Hence number of outcomes favorable to event A = n{A} = 5

Since n{S} = 36, Probability of event A = P{A} = n{A}/n{S} = 5/36

Now, P{A’} = 1 − P{A} = 1 − 5/36 = 31/36.

Thus odds against getting a sum of 6 = P{A’}/P{A} = 31:5.

问题36.获得以下优势的几率是多少?

(i)如果该牌是从一副洗牌良好的牌组中抽出的，则是黑桃?

解决方案：

Let W be the event of getting a spade from a well shuffled deck of cards and W’ be the event of not getting a spade.

We know there are 13 spade cards in a pack of cards.

Hence number of outcomes favorable to W = n{W} = 13

Probability of getting a spade = P{W} = n{W}/n{S} = 13/52 =1/4

Thus, probability of not getting a spade = P{W’} = 1 − P{W} = 1 −1/4 = 3/4

Odds in favour of getting a spade = P{W}/P{W’} = (1/4)/(3/4) = 1:4

(ii)国王?

解决方案：

Let K be the event of getting a king from a well shuffled deck of cards and K’ be the event of not getting a king.

We know there are 4 kings in a pack of cards.

Hence number of outcomes favorable to K = n{K} = 13

Probability of getting a king = P{K} = n{K}/n{S} = 4/52 =1/13

Thus, probability of not getting a king = P{K’} = 1 − P{K} = 1 −1/13 = 12/13

Odds in favour of getting a king = P{K}/P{K’} = (1/13)/(12/13) = 1:12

问题37.一个盒子装有10个红色大理石，20个蓝色大理石和30个绿色大理石。随机绘制5个弹珠。找到以下可能性：

(i)都是蓝色的

解决方案：

Clearly the box has 10 + 20 + 30 = 60 marbles and since 5 marbles have been drawn at random, number of outcomes in sample space = n{S} = ⁶⁰C₅

Let B be the event that all 5 marbles are blue.

Hence number of outcomes favorable to event B = n{B} = ²⁰C₅

Probability of B = n{B}/n{S} = ²⁰C₅ / ⁶⁰C₅ = 34/11977

Hence the probability of all 5 marbles being blue is 34/11977.

(ii)至少有一个是绿色的

解决方案：

Let G be the event that at least one marble is drawn.

Probability of G = 1 − Probability of drawing no green marble = 1 − ³⁰C₅/⁶⁰C₅ = 1 − 117/4484 = 4367/4484

Hence the probability of drawing at least one green marble is 4367/4484.

问题38：一个盒子中有6个红色大理石，编号从1到6，有4个白色大理石，编号从12到15。求出绘制大理石的概率为：

(i)白色

解决方案：

Since 1 marble has been drawn from a box of 10 marbles, number of outcomes in sample space = n{S} = ¹⁰C₁ = 10.

Let W be the event that the given marble is white.

Number of outcomes favorable to event W = n{W} = ⁴C₁ = 4.

Probability of W = n{W}/n{S} = 4/10 = 2/5

Hence the probability of the marble being white is 2/5.

(ii)白色和奇数

解决方案：

Let Q be the event that the marble drawn is white and odd numbered. We know there are 2 odd numbers between 12 and 15, (i.e., 13 and 15).

Number of outcomes favorable to event Q = 2

Since n{S} = 10, Probability of Q = n{Q}/n{S} = 2/10 = 1/5

Hence the probability of the marble being white and odd numbered is 1/5.

(iii)偶数

解决方案：

Let V be the event that the marble drawn is even numbered.

V = {2, 4, 6, 12, 14}

Number of outcomes favorable to event V = 5

Since n{S} = 10, Probability of V = n{V}/n{S} = 5/10 = 1/2

Hence the probability of the marble being even numbered is 1/2.

(iv)红色或偶数

解决方案：

Let G₁ be the event of getting a red marble. We know there are 6 red marbles, hence probability of G₁ = P{G₁} = 6/10

Let G₂ be the event of getting an even numbered marble. We know there are 5 even numbered marbles, hence probability of G₂ = P{G₂} = 5/10

Now, {G₁ ∩ G₂} = even numbered red marble = {2, 4, 6}

Thus, n{G₁ ∩ G₂} = 3 and Probability of {G₁ ∩ G₂} = P{G₁ ∩ G₂} = 3/10.

Applying the law of addition,

P{G₁ ∪ G₂} = P{G₁} + P{G₂} − P{G₁ ∩ G₂} = 6/10 + 5/10 − 3/10 = 8/10 = 4/5

Hence the probability of getting a red or even numbered marble is 4/5.

问题39.一个班级由10个男孩和8个女孩组成。随机选择三个学生。所选组具有以下几率的可能性：

(i)所有男孩

解决方案：

Since 3 students have been selected at random from a group of 18 students, number of outcomes in sample space = n{S} = ¹⁸C₃ = 816

Let B be the event that the selected group has all boys.

Hence number of outcomes favorable to B = n{B} =¹⁰C₃ = 120

Probability of event B = n{B}/n{S} = 120/816 = 5/34

Hence the probability that the group has all boys is 5/34.

(ii)所有女孩

解决方案：

Let G be the event that the selected group has all girls.

Hence number of outcomes favorable to G = n{G} = ⁸C₃ = 56

Since n{S} = 816, Probability of event G = n{G}/n{S} = 56/816 = 7/102

Hence the probability that the group has all girls is 7/102.

(iii)1个男孩和2个女孩

解决方案：

Let H be the event that the selected group has 1 boy and 2 girls.

Hence number of outcomes favorable to H = n{H} = ¹⁰C₁ × ⁸C₂ = 280

Since n{S} = 816, Probability of event H = n{H}/n{S} = 280/816 = 35/102

Hence the probability that the group has 1boy and 2 girls is 35/102.

(iv)至少1个女孩

解决方案：

Let C be the event that the selected group has at least 1 girl.

Probability of having at least 1 girl

=> 1 − P{no girl in the group} = 1 − P{all boys} = 1 − ¹⁰C₃/¹⁸C₃ = 1 − 5/34 = 29/34

Hence probability of having at least 1 girl is 29/34.

(v)最多1个女孩

解决方案：

Let T be the event that the selected group has at most 1 girl.

T = {0 girl, 1 girl}

Thus, n{T} = ⁸C₀ × ¹⁰C₃ + ⁸C₁ × ¹⁰C₂ = 480

Since n{S} = 816, Probability of event T = n{T}/n{S} = 480/816 =10/17

Hence probability of having at most 1 girl is 10/17.

问题40.从一张经过精心组合的52张纸牌中抽出5张纸牌。找出所有5张牌都是红心的概率。

解决方案：

Since 5 cards have been drawn from a well shuffled pack of 52 cards, number of outcomes in sample space = n{S} = ⁵²C₅ = 2598960.

Let H be the event that all 5 cards drawn are hearts. We know there are 13 hearts in a pack of 52 cards.

Hence number of outcome favorable to event H = n{H} = ¹³C₅ = 1287.

Probability of event H = P{H} = n{H}/n{S} = 1287/2598960 = 33/66640

Hence the probability of getting 5 hearts is 33/66640.

问题41.一个袋子里装着1至20的票。找到以下可能性：

(i)两张票上都有素数

解决方案：

Since 2 tickets have been drawn from a bag containing 20 tickets, number of outcomes in sample space = n{S} = ²⁰C₂ = 190.

Let N be the event that both the tickets have prime numbers on them. We know there are 8 prime numbers between 1 to 20, (i.e., 2, 3, 5, 7, 9, 11, 13, 17, 19).

Hence number of outcomes favorable to event N = ⁸C₂ = 28

Probability of event N = P{N} = n{N}/n{S} = 28/190 = 14/95

Hence the probability of both the tickets having prime numbers on them is 14/95.

(ii)一个有一个质数，另一个有4的倍数

解决方案：

Let T be the event that on one ticket there is a prime number and on the other there is a multiple of 4.

We know there are 8 prime numbers 5 multiples of 4 from 1 to 20.

Hence number of outcomes in event T = n{T} = ⁸C₁ × ⁵C₁ = 8 × 5 = 40.

Since n{S} = 190, Probability of event T = P{T} = 40/190 = 4/19

Thus the probability of getting a prime number on one ticket and a multiple of 4 on the other is 4/19.

问题42.包含7个白色，5个黑色和3个红色的球。随机抽出两个球。找到以下可能性：

(i)两个球都是红色的

解决方案：

Clearly the urn has 7 + 5 + 3 = 15 balls and since 2 balls have been drawn at random, number of outcomes in sample space = n{S} = ¹⁵C₂ =105.

Let R be the event that the two balls chosen are red.

Hence number of outcomes in event R = n{R} = ³C₂ = 3

Probability of R = n{R}/n{S} = 3/105 = 1/35.

Hence the probability of both balls being red is 1/35.

(ii)一个球是红色的，另一个是黑色的

解决方案：

Let X be the event that one ball is red and the other is black.

Hence number of outcomes in event X = n{X} = ³C₁ × ⁵C₁ = 15

Since n{S} = 105, Probability of X = n{X}/n{S} = 15/105 = 1/7.

Hence the probability of one ball being red and the other being black is 1/7.

(iii)一个球是白色的

解决方案：

Let D be the event that one ball is white. It means that the other ball can be from the 8 remaining balls, excluding the white balls.

Hence number of outcomes in event D = n{D} = ⁷C₁ × ⁸C₁ = 56

Since n{S} = 105, Probability of D = n{D}/n{S} = 56/105 = 8/15.

Hence the probability of one ball being white is 8/15.

问题43：A和B扔了一对骰子。如果A抛出9，则发现B抛出更高数字的机会。

解决方案：

Since a pair of dice has been thrown, number of items in sample space = n{S}= 62 = 36.

Let E be the event that B throws a higher number than 9, which is either 10 or 11 or 12.

Hence, E = {(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)}

Hence number of outcomes favorable to E = 6

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

Hence the probability of B throwing a higher number is 1/6.

问题44.在惠斯德(Whist)手中，指定玩家将4个国王保住的概率是多少?

解决方案：

We know that in a hand at Whist, each given player has a pack of 13 cards.

Hence number of outcomes in sample space = n{S} = ⁵²C₁₃

Let K denote the event that a player has 4 kings. It implies that the player must have 9 cards from the pack of the rest of 48 cards(excluding the 4 kings already with the player.)

Number of outcomes favorable to event K = n{K} = ⁴C₄ × ⁴⁸C₉ = 4 × ⁴⁸C₉

Probability of K = P{K} = n{K}/n{S} = 4 × ⁴⁸C₉ / ⁵²C₁₃ = 11/4165

Hence the probability that a specific player has all the 4 hearts is 11/4165.

问题45.求出一个概率，即在随机排列单词“ UNIVERSITY”的字母时，两个I在一起的可能性不大。

解决方案：

We know there are 10 letters in the word ‘UNIVERSITY’. Hence number of outcomes in sample space = n{S} = 10!

Let W denote the event that both I’s do not come together and let W’ denote the event that both I’s come together.

Hence number of outcomes favorable to event W’ = n{W’} = 2 × 9!

Thus, Probability of W’ = P{W’} = n{W’}/n{S} = 2 × 9!/10! = 2/10 = 1/5

Now, P{W} = 1 − P{W’} = 1 − 1/5 = 4/5

Hence the probability of not getting the two I’s together is 4/5.