第 12 类 RD Sharma 解决方案 - 第 31 章概率 - 练习 31.4 |设置 2
问题 13. 一个无偏的骰子被掷了两次。求第一次投掷得到 4、5 或 6 以及第二次投掷得到 1、2、3 或 4 的概率。
解决方案:
According to question:
Given an unbiased die is tossed twice.
Let us consider the events,
A = Getting 4, 5 or 6 on the first toss
B = 1, 2, 3 or 4 on the second toss
⇒ P(A) = 3 / 6 = 1/2
And, P(B) = 4/6 = 2/3
Now, we have to find that
P(Getting 4, 5 or 6 on the first toss and 1, 2, 3 or 4 on the second toss)
= P(A ∩ B) = P(A) P(B)
= 1/2 × 2/3 = 2/6 = 1/3
Hence, The required probability = 1/3
问题 14. 一个袋子里有 3 个红球和 2 个黑球。从中随机抽取一个球。它的颜色被记录下来,然后放回袋子里。进行第二次抽签并重复相同的程序。找出抽到 (i) 两个红球的概率
(ii) 两个黑球, (iii) 第一个红球和第二个黑球。
解决方案:
According to question:
It is given that,
A bag contain 3 red and 2 black balls.
Let us consider the following events,
A = Getting on red ball.
B = Getting one black ball.
Now, P(A) = 3/5 and, P(B) = 2/5
Now, We have to find that,
(i) P(Getting two red balls)
= P(A) P(A)
= 3/5 × 3/5 = 9/ 25
So, P(Getting two red balls) = 9/25
(ii) P(Getting two black balls)
= P(B) P(B)
= 2/5 × 2/5 = 4/ 25
So, P(Getting two black balls) = 4/25
(iii) P(Getting first red ball and second black ball)
= P(A) P(B)
= 3/5 × 2/5 = 6/ 25
So, P(Getting first red ball and second black ball) = 6/25
问题 15. 从一副洗好的牌中抽出三张牌并替换。找出牌是国王、王后和杰克的概率。
解决方案:
According to question:
Three cards are drawn with replacement. We know that there
are 52 cards in which 4 kings, 4 queens and 4 jacks.
Let us consider the following events
A = Drawing a king
B = Drawing a queen
C = Drawing a jack
Now, P(A) = 4/ 52 = 1/13
⇒ P(B) = 4/52 = 1/13
⇒ P(C) = 4/52 = 1/13
Now we have to find that,
P(Cards drawn are king, queen and jack)
Since drawing order are different,
= P(A ∩ B ∩ C) + P(A ∩ C ∩ B) + P(B ∩ A ∩ C) +
P(B ∩ C ∩ A) + P(C ∩ A ∩ B) + P(C ∩ B ∩ A)
= P(A) P(B) P(C) + P(A) P(C) P(B) + P(B) P(A) P(C) +
P(B) P(C) P(A) + P(C) P(A) P(B) +P(C) P(B) P(A)
Now, putting the values of P(A), P(B) and P(C) in the above expression, then we get
= 1/13 × 1/13 × 1/13 + 1/13 × 1/13 × 1/13 + 1/13 × 1/13 × 1/13 +
1/13 × 1/13 × 1/13 + 1/13 × 1/13 × 1/13 + 1/13 × 1/13 × 1/13
= (1/13 × 1/13 × 1/13) × 6
= 6/2197
Hence, The required probability = 6/2197
问题 16:某公司制造的工件由 X 和 Y 两部分组成。在 X 部分的制造过程中,100 个零件中有 9 个可能有缺陷。同样,在零件 Y 的制造过程中,100 个中有 5 个可能有缺陷。计算组装产品不会有缺陷的概率。
解决方案:
According to question:
It is given that,
Part X has 9 out of 100 defective
So, it is clear that Part X has 91 out of 100 non-defective
Now, Part Y has 5 out of 100 defective
Here, Part Y has 95 out of 100 non-defective
Now let us consider the events,
X = A non-defective part X
Y = A non-defective part Y
So, P(X) = 91/100 and P(Y) = 95/100
We have to find the probability of assembled product will not be defective.
Now, P(Assembled product will not be defective)
= P(Neither X defective nor Y defective)
= P(X ∩ Y) = P(X) P(Y)
= 91/100 × 95/100
= 0.8645
Hence, The required probability = 0.8645
问题 17. A 击中目标的概率是 1/3,B 击中目标的概率是 2/5。如果 A 和 B 都向目标射击,目标被击中的概率是多少?
解决方案:
According to question,
Given that,
Probability that A hits a target = 1/3
P(A) = 1/3
And, Probability that B hits a target = 2/5
P(B) = 2/5
We have to find that,
P(Target will be hit)
= 1 – P(target will not be hit)
= 1 – P(Neither A nor B hits the target)
= 1 – P(A‘ ∩ B‘)
= 1 – P(A‘) P(B‘) = 1 – [1 – P(A)] [1 – P(B)]
= 1 – [1 – 1/3 ] [1 – 2/5]
= 1 – 2/3 × 3/5 = 1 – 2/5
= 3/5
Hence, The required probability = 3/5
问题 18. 高射炮最多可以对远离它的敌机进行 4 次射击。第一次、第二次、第三次和第四次击中飞机的概率分别为 0.4、0.3、0.2 和 0.1。枪击中飞机的概率是多少?
解决方案:
According to question:
Given that,
An anti-aircraft gun can take a maximum 4 shots at an enemy plane.
Let us consider the following events,
A = Hitting the plane at first shot
B = Hitting the plane at second shot
C = Hitting the plane at third shot
D = Hitting the plane at fourth shot
⇒ P(A) = 0.4, P(B) = 0.3, P(C) = 0.2, P(D) = 0.1
We have to find that
P (Gun hits the plane)
= 1 – P(None of four shots hit the plane)
= 1 – P(A‘ ∩ B‘ ∩ C‘ ∩ D‘)
= 1 – P(A‘) P(B‘) P(C‘) P(D‘)
= 1 – [1 – P(A)] [1 – P(B)] [1 – P(C)] [1 – P(D)]
Now, putting the value of P(A), P(B), P(C) And P(D) in the above expression,
= 1 – (0.6) (0.7) (0.8) (0.9)
= 1 – 0.3024
= 0.6976
Hence, The required probability = 0.6976
问题 19. 反对某一事件的概率是 5 比 2,支持另一事件的概率是 6 比 5,独立于前者。求 (a) 至少有一个事件发生的概率,以及) 不会发生任何事件。
解决方案:
According to question:
It is given that,
Let us consider the following events, A and B
The odd against a certain event (Say A) are 5 to 2
⇒ P(A‘) = 5/(5 + 2) = 5/7
And, The odd in favor of another event (Say B) are 6 to5
⇒ P(B) = 6/(5 + 6) = 6/11
⇒ P(B‘) =1 – 6/11 = 5/11
Now,
(a) P(At least one of the events will occur)
= 1 – P(None of the events occur)
= 1 – P(A‘∩B‘) = 1 – P(A‘) P(B‘)
= 1 – 5/7 × 5/11 = 1 – 25/77
= 52/77
Hence, the required probability = 52/77
(b) P(None of the events will occur)
= P(A‘ ∩ B‘) = P(A‘) P(B‘)
= 5/7 × 5/11 = 25/77
Hence, the required probability = 25/77
问题 20. 骰子被掷三次。求至少一次得到奇数的概率。
解决方案:
According to question:
Given that, A die is thrown thrice
Let us consider the following events,
A = Getting an odd number in a throw of die
Since, 1, 3, 5 are odd numbers on die So,
P(A) = 3/6 = 1/2
P(A‘) = 1/2
We have to find that,
P(Getting an odd number at least once)
= 1 – P(Getting no odd number)
= 1 – P(A‘) P(A‘) P(A‘)
= 1 – 1/2×1/2×1/2 = 1 – 1/8
= 7/8
Hence, the required probability = 7/8
问题 21. 从一个包含 10 个黑球和 8 个红球的盒子中随机抽取两个球并替换。找出概率
(i) 两个球都是红色的
(ii) 第一个球是黑色的,第二个是红色的。
(iii) 其中一个是黑色的,另一个是红色的
解决方案:
According to question:
It is given that,
The box contains 10 black balls and 8 red balls.
Then, P(black ball) = 10/18
P(red ball) = 8/18
(i) P(Both balls are red) = 8/18 × 8/18
= 4/9 × 4/9 = 16/81
(ii) P(First ball is black and second is red.)
= 10/18 × 8/18 = 20/81
(iii) P(One of them is black and other is red)
= 10/18 × 8/18 + 8/18 × 10/18
= 2 ×(20/81) = 40/81
问题 22. 一个瓮中有 4 个红球和 7 个蓝球。随机抽取两个球并替换。求得到 (i) 2 个红球 (ii) 2 个蓝球 (iii) 一个红球和一个蓝球的概率。
解决方案:
According to question:
It is given that,
Urn contains 4 red and 7 blue balls. Two balls are drawn at random with replacement.
Let us consider the events,
A = Getting one red ball from the urn.
P(A) = 4/11
B = Getting one blue ball from urn.
P(B) = 7/11
Now, we have to find that,
(i) P(Getting 2 red balls)
= P(A).P(A)
= 4/11×4/11 = 16/121
Hence, the required probability = 16/121
(ii) P(Getting 2 blue balls)
= P(B).P(B)
= 7/11 × 7/11 = 49/121
Hence, the required probability = 49/121
(iii) P(Getting one red and one blue ball)
= P(A).P(B) + P(B).P(A)
= 4/11×7/11 + 7/11× 4/11
= 28/121 + 28/121 = 56/121
Hence, the required probability = 56/121
问题 23. 两个学生 A 和 B 准时到校的概率分别是 3/7 和 5/7。假设“A 准时到来”和“B 准时到来”这两个事件是独立的,求其中只有一个准时到校的概率。至少写出及时到校的一项优势。
解决方案:
According to question:
It is given that,
The events, ” A coming in time” and ” B coming in time” are independent.
Let us consider the events,
A = A coming in time
B = B coming in time
Now, P(A) = 3/7, and P(A‘) = 1- 3/7 = 4/7
P(B) = 5/7, and P(B‘) = 1- 5/7 = 2/7
Now, we have to find that,
P(Only one coming in time)
⇒ P(A ∩ B‘) + P(A‘ ∩ B) = P(A) × P(B‘) + P(A‘) × P(B) -(Since, A and B are independent events)
⇒ 3/7 × 2/7 + 4/7 × 5/7 = 6/49 + 20/49 = 26/49
Hence, The required probability = 26/49
The advantages of coming school in time is that you have not to
give the late fine, you will not miss any part of the lecture and many more.
问题 24:将两个骰子放在一起,记下总分。事件 E、F 和 G 分别是“总数为 4”、“总数为 9 或更多”和“总数可被 5 整除”。计算 P(E)、P(F) 和 P(G) 并确定哪些事件对(如果有)是独立的。
解决方案:
According to question:
It is given that,
Two dice are thrown together hence, the total sample space will be 36.
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
So, n(S) = 36
Now, let us consider the events,
E be the event of getting a total of 4
E = {(1, 3), (3, 1), (2, 2)}
n(E) = 3
Now, P(E) = n(E)/n(S) = 3/36 = 1/12
Now, F be the event of getting a total 9 or more.
F = {(3, 6), (6, 3), (4, 5), (5, 4), (4, 6), (6, 4), (5, 5), (5, 6), (6, 5), (6, 6)}
n(F) = 10
P(F) = n(F)/n(S) = 10/36 = 5/18
Now, G be the event of getting a total divisible by 5.
G = {(1, 4), (4, 1), (2, 3), (3, 2), (4, 6), (6, 4), (5, 5)}
n(G) = 7
P(G) = n(G) / n(S) = 7/36
Hence, we can see that no pair is independent.
问题 25. 假设 A 和 B 是两个独立事件,使得 P(A) = p1 和 P(B) = p2。用文字描述其概率为:
(i) p1p2 (ii) (1 – p1)p2 (iii) 1 – (1 – p1)(1 – p2) (iv) p1 + p2 = 2p1p2
解决方案:
According to question:
The events are said to be independent, if occurrence or non-occurrence of one
doesn’t affect the probability of the occurrence or non-occurrence of the other.
Now,
(i) p1p2 = P(A).P(B)
⇒ Both A and B occur.
(ii) (1 – p1)p2 = (1 – P(A)) P(B) = P(A‘) P(B)
⇒ Event A doesn’t occur but B occurs.
(iii) 1 – (1 – p1)(1 – p2) = [1 – (1 – P(A))(1 – P(B))] = (1 – P(A‘) P(B‘))
⇒ At least one of the events A or B occurs.
(iv) p1 + p2 = 2p1p2
⇒ P(A) + P(B) = 2 P(A) P(B)
⇒ P(A) + P(B) – 2 P(A) P(B) = 0
⇒ P(A) – P(A) P(B) + P(B) – P(A) P(B) = 0
⇒ P(A) – P(A) P(B) + P(B) – P(A) P(B) = 0
⇒ P(A) – P(A) P(B) + P(B) – P(A) P(B) = 0
⇒ P(A) (1- P(B)) + P(B) (1 – P(A)) = 0
⇒ P(A) P(B‘) + P(B) P(A‘) = 0
⇒ P(A) P(B‘) + P(B) P(A‘) = 0,
⇒ P(A) P(B‘) = – P(B) P(A‘)
⇒ Exactly one of A and B occurs.