问题1：从一副52张卡片中，一张一张一张地抽两张，不用补发。找出他们俩都是国王的可能性。
解决方案：
Let the desired events be,
A = first card is king
B = second card is also king
and probability of the event A is, P(A).
P(A) = 4/52 = 1/13 [There are 4 kings in the 52 cards set]
As the king is withdrawn without replacement there are three kings left only.
And the number of cards left is 51 as well.
Hence,
P(B/A) = 3/51 = 1/17
Hence, the required probability,
P(A∩B) = P(A) × P(B/A) = (1/13) × (1/17)
or, P(A∩B) = 1/221 (ans)
问题2：从52张纸牌中，一张一张一张地抽出四张，不用补发。找出所有这些人都是A的可能性。
解决方案：
Let the desired events be,
A = an ace in the first draw
B = an ace in the second draw
C = an ace in the third draw
D = an ace in the fourth draw
The probability of the event x is P(x)
P(A) = 4/52 = 1/13 [There are 4 aces and 52 cards remaining]
P(B/A) = 3/51 = 1/17 [There are 3 aces and 51 cards remaining]
P(C/A∩B) = 2/50 = 1/25 [There are 2 aces and 50 cards remaining]
P(D/A∩B∩C) = 1/49 [There is 1 ace and 49 cards remaining]
The required probability,
P(A∩B∩C∩D) = P(A) × P(B/A) × P(C/A∩B) × P(D/A∩B∩C)
=(1/13) × (1/17) × (1/25) × (1/49)
=1/270725 (ans)
问题3：找到从装有5个红色和7个白色小球的袋子中依次抽出2个白色小球的机会。先拉的球永远不会被替换。
解决方案：
Let the desired events be,
A = White ball at first draw
B = White ball at second draw
The probability of the event x is P(x).
P(A) = 7/12 [Any of the seven white balls from the 12-ball-set]
P(B/A) = 6/11 [6 white balls remaining and 11 balls left]
The required probability,
P(A∩B) = P(A) × P(B/A)
= (7/12) × (6/11) = 7/22 (ans)
问题4：一个袋子包含25张票，从1到25编号。抽出一张票，然后抽出另一张不更换票。找出两张票证均显示偶数的可能性。
解决方案：
Let the desired events be,
A = Even number at first draw
B = Even number at second draw
The probability of the event x is P(x).
P(A) = 12/25 [Any of the12 even numbers less than 25]
P(B/A) = 11/24 [Any of the 11 remaining even number tickets out of 24]
The required probability,
P(A∩B) = P(A) × P(B/A)
= (12/25) × (11/24) = 11/50 (ans)
问题5：从一副纸牌中，一张一张一张地抽出三张牌，无需更换。找出它是一张黑桃纸牌的可能性。
解决方案：
Let the desired events be,
A = a spade in the first draw
B = a spade in the second draw.
C = a spade in the third draw.
P(A) = 13/52 = 1/4 [There are 13 spades and 52 cards remaining]
P(B/A) = 12/51 = 4/17 [There are 12 spades and 51 cards remaining]
P(C/A∩B) = 11/50 [There are 11 spades and 50 cards remaining]
The required probability,
P(A∩B∩C) = P(A) × P(B/A) × P(C/A∩B)
=(1/4) × (4/17) × (11/50) = 11/850 (ans)
问题6：
(i)从52张纸牌中抽出两张纸牌而无需更换。找出双方都是国王的可能性。
解决方案：
Let the desired events be,
A = first card is king
B = second card is also king
and probability of the event A is, P(A).
P(A) = 4/52 = 1/13 [There are 4 kings in the 52 cards set]
As the king is withdrawn without replacement there are three kings left only.
And the number of cards left is 51 as well.
Hence,
P(B/A) = 3/51 = 1/17
Hence, the required probability,
P(A∩B) = P(A) × P(B/A) = (1/13) × (1/17)
or, P(A∩B) = 1/221 (ans)
(ii)从52张纸牌中抽出两张纸牌而无需更换。找出第一个为王，第二个为王牌的概率。
解决方案：
Let the desired events be,
A = first card is king
B = second card is also king
and probability of the event A is, P(A).
P(A) = 4/52 = 1/13 [There are 4 kings in the 52 cards set]
As the king is withdrawn without replacement the number of cards left is 51 and it has 4 aces.
Hence,
P(B/A) = 4/51 [4 aces in 51 cards set]
Hence, the required probability,
P(A∩B) = P(A) × P(B/A) = (1/13) × (4/51)
or, P(A∩B) = 4/663 (ans)
(iii)从52张纸牌中抽出两张纸牌而无需更换。找到第一个是心脏，第二个是红色的概率。
解决方案：
There are 13 heart cards and 26 red cards, if we withdraw a heart card 25 red cards remain.
Let the desired events be,
A = first card is heart.
B = second card is red.
The probability of the event x is P(x).
P(A) = 13/52 = 1/4 [There are 13 hearts in the 52 cards set]
If we withdraw a heart card 25 red cards remain
P(B/A) = 25/51 [26 red cards in 51 cards set]
Hence, the required probability,
P(A∩B) = P(A) X P(B/A) = (1/4) X (25/51)
or, P(A∩B) = 25/204 (ans)
问题7：一个袋子包含20张票，编号从1到20。先抽出一张票，然后再抽出另一张不用更换的票。找到第一个显示偶数，第二个显示奇数的概率。
解决方案：
Let the desired events be,
A = Even number at first draw
B = Odd number at second draw
The probability of the event x is P(x).
P(A) = 10/20 = 1/2 [Any of the 10 even numbers less than 20]
P(B/A) = 10/19 [Any of the 10 remaining odd number tickets out of 19]
The required probability,
P(A∩B) = P(A) × P(B/A)
= (1/2) × (10/19) = 5/19 (ans)
问题8：包含3个白色，4个红色和5个黑色的球。两个球一张一张地抽出来，不用替换。至少一个球是黑色的概率是多少?
解决方案：
Let the desired events be,
A = Black ball at first draw
B = Black at second draw.
Complement of desired events be,
A’ = A white ball or red ball in first draw.
B’ = A white ball or red ball in second draw.
P(A’) = 7/12 [7 non-black balls out of 12]
P(B’/A’) = 6/11 [6 non-black balls out of 11]
Hence, the required probability,
P(At least one ball is black)
= P(A U B)
= 1 – P(A U B)’
=1 – P(A’ ∩ B’)
=1 – P(A’)P(B/A)’
=1 – (7/12 × 6/11)
=15/22 (ans)
问题9：一个袋子包含5个白色的7个红色的球和3个黑色的球。如果三个球被一一抽出而不更换。找出它们都不是红色的可能性。
解决方案：
Let the desired events be,
A = No red ball in first draw.
B = No red ball in second draw.
C = No red ball in the third draw.
P(A) = 8/15 [8 non-red balls of 15]
P(B/A) = 7/14 = 1/2 [7 non-red balls of 14]
P(C/A∩B) = 6/13 [6 non-red balls of 13]
The required probability,
P(A∩B∩C) = P(A) × P(B/A) × P(C/A∩B)
= 8/15 × 1/2 × 6/13
= 8/65 (ans)
问题10：从一副精心包装的52张纸牌中抽出一张纸牌，然后抽出第二张纸牌。找到第一张牌是心脏，第二张牌是菱形的概率。如果未更换第一张卡。
解决方案：
Let the desired events be,
A = first card is heart
B = second card is diamond
c = 13/52 = 1/4 [13 hearts out of 52 cards]
P(B/A) = 13/51 [13 diamonds out of 51 cards]
The required probability,
P(A∩B)
= P(A) × P(B/A)
=1/4 × 13/51
=13/204 (ans)
问题11：包含10个黑球和5个白球。从the中抽出两个球，一个接一个地替换，而无需更换。两个抽出的球都是黑色的概率是多少?
解决方案：
Let the desired events be,
A = first ball is black
B = second ball is black
P(A) = 10/15 = 2/3 [10 black balls of 15]
P(B/A) = 9/14 [9 black balls of 14]
The required probability,
P(A∩B)
= P(A)P(B/A)
= 2/3 × 9/14
= 3/7 (ans)
问题12：连续抽出三张纸牌，而没有从一包52张洗净的纸牌中更换。前两张牌为国王，第三张牌为王牌的概率是多少?
解决方案：
Let the desired events be,
A = first card is king
B = second card is king
C = third card is an ace
P(A) = 4/52 = 1/13 [There are 4 kings and 52 cards remaining]
P(B/A) = 3/51 = 1/17 [There are 3 kings and 51 cards remaining]
P(C/A∩B) = 4/50 = 2/25 [There are 4 aces and 50 cards remaining]
The required probability,
P(A∩B∩C)
= P(A)P(B/A)P(C/A∩B)
= 1/13 × 1/17 × 2/25
= 2/5525 (ans)
问题13：一盒橘子由三个随机选择的橘子检查而没有更换，它们被检查。如果所有三个橙子都很好，则表示该包装盒可以出售，否则将被拒绝。确定包含15个橙子的盒子(其中12个好和3个坏的盒子)将被批准出售的可能性。
解决方案：
Let the desired events be,
A = first orange is good.
B = second orange is good.
C = third orange is good.
P(A) = 12/15 = 4/5 [There are 12 good oranges among 15]
P(B/A) = 11/14 [There are 11 good oranges among 14]
P(C/A∩B) = 10/13 [There are 10 good oranges among 13]
The required probability,
P(A∩B∩C)
= P(A)P(B/A)P(C/A∩B)
= 4/5 × 11/14 × 10/13
= 44/91 (ans)
问题14：一个袋子包含4个白色，7个黑色和5个红色的球。三个球被一个接一个地拉出而无需更换。找出画出的球分别是白色，黑色和红色的概率。
解决方案：
Let the desired events be,
A = first ball is white.
B = second ball is black.
C = third ball is red.
P(A) = 4/16 = 1/4 [There are 4 white balls among 16]
P(B/A) = 7/15 [There are 7 black balls among 15]
P(C/A∩B) = 5/14 [There are 5 red balls among 14]
The required probability,
P(A∩B∩C)
= P(A)P(B/A)P(C/A∩B)
= 1/4 × 7/15 × 5/14
= 1/24 (ans)