问题16.一个袋子包含7个白色,5个黑色和4个红色的球。如果随机抽出两个球,则求出以下概率:
(i)两个球都是白色的
解决方案:
There are 7+5+4 =16 balls in the bag and since 2 balls have been drawn at random, the total number of outcomes in sample space= n{S}= 16C2 = 120.
Let W be the event of getting both white balls.
Since there are 7 white balls, number of outcomes in W= n{W} = 7C2 = 21.
Probability of event= P{W} = n{W}/n{S} = 21/120 = 7/40
Hence probability of both balls being white is 7/40.
(ii)一个球是黑色的,另一个是红色的
解决方案:
Let A be the event of getting one black and one red ball.
Hence number of outcomes in A= n{A}= 5C1 × 4C1 = 20.
Since n{S}= 120, Probability of A= P{A}= n{A}/n{S}= 20/120= 1/6
Hence probability of one ball being black and the other being red is 1/6.
(iii)两个球的颜色相同
解决方案:
Let O be the event of getting both balls of same color.
Hence number of outcomes in O= n{O}= 7C2 +5C2 + 4C2 = 37.
Since n{S}= 120, Probability of O= P{O}= n{O}/n{S}= 37 / 120.
Hence probability of both the balls being of the same color is 37/120.
问题17.一个袋子包含6个红色,4个白色和8个蓝色的球。如果随机抽出三个球,则求出以下概率:
(i)一个是红色,两个是白色
解决方案:
There are 6+4+8 = 18 balls in the bag and since 3 balls have been drawn at random, the total number of outcomes in sample space= n{S}= 18C3 = 816.
Let Y be the event of getting one red and two white balls.
Hence number of outcomes in Y = n{Y} = 6C1 × 4C2 = 36.
Probability of event= P{Y} = n{Y}/n{S} = 36/816 = 3/68
Hence probability of one ball being red and the other two being white is 3/68.
(ii)两个是蓝色,一个是红色
解决方案:
Let L be the event of getting two blue and one red balls.
Hence number of outcomes in L= n{L}= 8C2 × 6C1=168.
Since n{S}= 120, Probability of L= P{L}= n{L}/n{S}= 168/816= 7/34
Hence probability of one ball being red and the other two being blue is 7/34.
(iii)一个是红色
解决方案:
Let X be the event that one of the balls must be red.
Hence number of outcomes in X= n{X}= 6C1 × 4C1 × 8C1 + 6C1 × 4C2 + 6C1 × 8C2 = 396.
Since n{S}= 120, Probability of X= P{X}= n{X}/n{S} = 396/816 = 33/68
Hence probability of one ball being red is 33/68.
问题18.从一包52张纸牌中抽取五张纸牌。这5个将包含的机会是:
(i)一张王牌
解决方案:
Since five cards are drawn at random from a pack of 52 cards, total number of outcomes in sample space= n{S}= 52C5 = 2598960
Let E be the event that exactly only one ace is present.
Hence number of outcomes in E= n{E}= 4C1 × 48C4 = 778320
Probability of event= P{E} = n{E}/n{S}= 778320/2598960= 3243/10829
Hence probability of getting just one ace is 3243/10829.
(ii)至少一个王牌
解决方案:
Let K be the event that at least one ace is present in the cards drawn from the pack of 52 cards.
Hence event K= {1 or 2 or 3 or 4 ace(s)}
Hence n{K} = 4C1 × 48C4 + 4C2 × 48C3 + 4C3 × 48C2 + 4C4 × 48C1 = 886656
Probability of event K= P{K} = n{K}/n{S}= 886656/2598960= 18472/54145
Hence probability of getting at least one ace is 18472/54145.
问题19.将面部卡从完整包装中取出。在剩余的40张卡中,随机抽取4张。他们属于不同西装的概率是多少?
解决方案:
Since there are 12 face cards(4 queens, 4 kings, 4 jacks and 4 aces) in a pack of 52 cards, the total number of cards left are 40.
Out of these 40 cards, the number of ways of choosing 4 cards= n{S}= 40C4 = 91390
Let E be the event that 4 cards belong to different suit.
Hence number of outcomes in E= n{E}= 10C1 × 10C1 × 10C1 × 10C1 = 10000
Probability of event E= P{E} = n{E}/n{S}= 10000/91390= 1000/9139
Hence the probability of the cards belonging to different suits is 1000/9139.
问题20.市议会有四男六女。如果随机选择一名理事会成员委员会成员,那么该成员是女性的可能性有多大?
解决方案:
There are four men and six women on the city councils. Hence total number of people on the council is 10.
Since 1 council member is selected at random number of outcomes in sample space= n{S}= 10C1 = 10.
Let E be the event that it is a woman.
Hence number of outcomes in E= n{E} = 6C1 = 6
Probability of event E= P{E} = n{E}/n{S} = 6/10 = 3/5
Hence the probability of the member chosen at random being a woman is 3/5.
问题21.一个盒子装有100个灯泡,其中20个有缺陷。选择了10个灯泡进行检查。找到以下可能性:
(i)全部10个有缺陷
解决方案:
Since ten bulbs have been drawn at random for inspection from a bag of 100 bulbs, total possible outcomes in sample space= n{S}= 100C10
Let E be the event that all ten bulbs are defective. We know 20 bulbs are defective.
Hence number of outcomes in E= n{E}= 20C10
Probability of event E= P{E} = n{E}/n{S}= 20C10 / 100C10
Hence the probability of all 10 bulbs being defective is 20C10 / 100C10.
(ii)全部10个都好
解决方案:
Let U be the event that all ten good bulbs are selected.
We know that 20 bulbs are defective. So, number of good bulbs= 100− 20= 80.
Since all 10 bulbs are good, number of outcomes in event U= n{U} = 80C10
Since n{S}= 100C10, Probability of event U=P{U}= n{U}/n{S}= 80C10 / 100C10
Hence the probability of all balls being good is 80C10 / 100C10.
(iii)至少一个有缺陷
解决方案:
Let R be the event that at least one bulb is defective. Since there are 10 defective bulbs, let us assign a number to each number from 1 to 10.
Thus, R= {1,2,3,4,5,6,7,8,9,10}
Let R′ be the event that none of the bulb is defective
n{R’}= 80C10
Since n{S}= 100C10 , Probability of R’= P{R’}= n{R’}/n{S}= 80C10 / 100C10
So, P{R}= 1 – P{R’}= 1 – 80C10 / 100C10
Hence the probability of getting at least one defective bulb is 1 – 80C10 / 100C10
(iv)没有缺陷
解决方案:
Let M be the event that none of the selected bulb is defective. We know there are 80 good bulbs or 80 non- defective bulbs in the bag.
Hence number of outcomes of M= n{M}= 80C10
Since n{S}= 100C10, Probability of M= P{M}= n{M}/n{S}= 80C10 / 100C10
Hence the probability of getting all good bulbs is 80C10 / 100C10.
问题22。找出“ SOCIAL”元音字母的字母随机排列在一起的可能性。
解决方案:
We know there are 6 alphabets in the word ‘SOCIAL’
Hence the number of ways of arranging the letters = n{S}= 6!= 720
Let E be the event that vowels come together.
The vowels in SOCIAL are A, I, O. So, number of vowels= 3
Thus, number of ways to arrange them where the three vowels come together= n{E}= 4! × 3!= 144
Probability of E= P{E}= n{E}/n{S}= 144/720= 1/5.
Hence the probability of getting all the three vowels together is 1/5.
问题23:“ CLIFTON”一词的字母随机排列。两个元音在一起的机会是多少?
解决方案:
We know there are 6 alphabets in the word ‘CLIFTON’
Hence the number of ways of arranging the letters = n{S}= 7!= 5040.
Let E be the event that vowels come together
The vowels in the word CLIFTON are I, O. So number of vowels= 3.
Thus, number of ways to arrange them where the three vowels come together= n{E}= 6! × 2!= 1440
Probability of E= P{E}= n{E}/n{S}= 1440/5040= 2/7.
Hence the probability of getting both vowels together is 2/7.
问题24.“ FORTUNATE”一词的字母随机排列。两个“ T”聚在一起的机会是什么?
解决方案:
We know there are 10 alphabets in the word ‘FORTUNATE’
Hence the number of ways of arranging the letters = n{S}= 10!
Let E be the event that two ‘T’ come together.
Thus, number of ways to arrange them where the three vowels come together= n{E}= 2 × 9!
Probability of E = P{E}= n{E}/n{S}= 2 × 9!/10!= 2/10= 1/5
Hence the probability of getting both ‘T’ together is 1/5.
问题25.从2名男性和2名女性中选出一个由2人组成的委员会。找出委员会将具有的可能性:
(i)没有人
解决方案:
Since 2 people are to be selected from a group of 4 people, number of outcomes in sample space= n{S}= 4C2 = 6
Let M be the event denoting no man gets chosen from the committee. Hence, both the women get chosen.
Number of outcomes in event M= n{M}= 2C2 = 1
Probability of event M= P{M}= n{M}/n{S}= 1/6
Hence the probability of selecting no men is 1/6.
(ii)一个人
解决方案:
Let O be the event that one man is in the committee.
Since there are 2 men, number of outcomes in event O= n{S}= 2C1 × 2C1 = 2 × 2 = 4
Since n{S}= 6, Probability of event O=P{O}= n{O}/n{S}= 4/6
Hence the probability of choosing one man is 4/6.
(iii)两名男子
解决方案:
Let T be the event that 2 men were chosen in the committee.
Since there are 2 men in the group, number of outcomes in event T= n{T}= 2C2 = 1
Since n{S}= 6, Probability of event T=P{T}= n{T}/n{S}=1/6
Hence the probability of choosing two men is 1/6.
问题26.如果赞成某事件的几率是2:3,请找到该事件发生的概率。
解决方案:
Let the given event be denoted by E.
Since the odds are in the favour of the event are in the ratio 2:3, number of outcomes in sample space=n{S}= 2a + 3a= 5a.
Also, number of outcomes in event= n{E}= 2a
Probability of event E= P{E}= 2a/5a= 2/5
Hence the probability of occurrence of event E is 2/5.
问题27.如果某事件的赔率是7:9,则查找未发生该事件的概率。
解决方案:
Since the odds against the event are 7:9, number of outcomes in the sample space= n{S}= 7a + 9a= 16a.
Let E be the event that the particular event takes place.
Hence the number of outcomes in event E= 9a
Probability of E= P{E}= n{E}/n{S}= 9a/16a= 9/16
Thus, probability of non-occurrence= P{E’}= 1 − P{E}= 1 − 9/16= 7/16
Hence the probability of non-occurrence of event is 7/16.
问题28.从一个装有2个白色,3个红色,5个绿色和4个黑色的球袋中随机抽取两个球,一个接一个地替换。找到两个球都具有不同颜色的可能性。
解决方案:
The bag contains 2 + 3 + 5 + 4= 14 balls, and since two balls have been drawn without replacement, the total number of outcomes in the sample space= n{S}= 14C2 = 91.
Let E be the event that all the balls are of different colors and let E’ be the event that both the balls chosen are of the same color.
Event E’= {WW, RR, GG, BB}
Number of outcomes in event E’ = 2C2 + 3C2 + 5C2 + 4C2 = 20.
Probability of E’ = P{E’} = n{E’}/n{S} = 20/91
Thus, P{E} = 1 − P{E’}= 1 − 20/91= 71/91
Hence the probability of both the balls being of different colors is 71/91.
问题29.掷出两个不偏不倚的骰子。找到以下可能性:
(i)既不会出现双打,也不会出现总共8个
解决方案:
Since two unbiased dice are thrown, the total number of outcomes in sample space= n{S}= 62 = 36.
Let E be the event that neither a doublet or a total of 8 will appear and E’ be the event that either a doublet or total of 8 will appear.
E’= {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6), (2,6), (3,5), (5,3), (6,2)}
n{E’} = 10
Probability of E’= P{E’}= n{E’}/n{S}= 10/36
Thus, P{E}= 1 − P{E’}= 1 − 10/36= 26/36 = 13/18
Hence the probability that neither a doublet or a total of 8 will appear is 13/18.
(ii)在两个骰子中获得的数字总和既不是2的倍数也不是3的倍数。
解决方案:
Let E be the event that the sum of the numbers obtained in the two dice is neither a multiple of 2 nor a multiple of 3 and E’ be the event that the sum of the numbers obtained in the two dice is either a multiple of 2 nor a multiple of 3.
E’= {(1,1), (1,2), (2,1), (1,3), (2,2), (3,1), (1,5), (2,4), (3,3), (4,2), (5,1), (2,6), (3,5), (4,4), (5,3), (6,2), (3,6), (4,5), (5,4), (6,3), (4,6), (5,5), (6,4), (6,6)}
n(E’) = 24
Since n{S}= 36, Probability of E’= P{E’}= n{E’}/n{S}= 24/36
Thus, P{E}= 1 − P{E’}= 1 − 24/36= 12/36 = 1/3
Hence the probability that the sum of the numbers obtained in the two dice is neither a multiple of 2 nor a multiple of 3 is 1/3.
问题30.一个袋子包含8个红色,3个白色和9个蓝色的球。如果随机抽出三个球,请确定
(i)所有三个球都是蓝色的球。
解决方案:
The bag contains 8 + 3 + 9 = 20 balls, and since three balls have been drawn at random, the total number of outcomes in the sample space= n{S} = 20C3 = 1140
Let E be the event that all the three balls are blue balls.
Number of outcomes in event E= 9C3 = 84.
Probability of E= P{E}= n{E}/n{S}= 84/1140 = 7/95
Hence the probability that all the three balls are blue balls is 7/95.
(ii)所有的球都是不同的颜色。
解决方案:
Let E be the event that all the balls are of different colors.
Number of outcomes in event E= 8C1 × 3C1 × 9C1 = 216.
Since n{S}= 1140, Probability of E= P{E}= n{E}/n{S}= 216/1140 = 18/95
Hence the probability that all the balls are of different colors is 18/95.