第 12 类 RD Sharma 解决方案 – 第 31 章概率 – 练习 31.5 |设置 3
问题 24. X 在考试中选择了数学、物理和化学科目。他在这些科目中获得 A 级的概率分别为 0.2、0.3 和 0.5。找出他得到的概率。
(i) 所有科目的 A 级 (ii) 没有科目的 A 级 (iii) 两门科目的 A 级。
解决方案:
According to question,
It is given that,
P(Getting grade A in mathematics), P(A) = 0.2 and, P(A‘) = 0.8
P(Getting grade A in physics), P(B) = 0.3 and, P(B‘) = 0.7
P(Getting grade A in chemistry), P(C) = 0.5 and, P(C‘) = 0.5
Now,
(i) P(Grade A in all subjects)
= P(A) × P(B) × P(C)
= 0.2 × 0.3 × 0.5
= 0.03
Hence, Required probability = 0.03
(ii) P(Grade A in no subject)
= P(A‘) × P(B‘) × P(C‘)
= 0.8 × 0.7 × 0.5
= 0.28
Required probability = 0.28
(iii) P(Grade A in two subjects)
= P(Not grade A in mathematics) + P(Not grade A in physics) + P(Not grade A in chemistry)
= P(A‘) × P(B) × P(C) + P(A) × P(B‘) × P(C) + P(A) × P(B) × P(C‘)
= 0.8 × 0.3 × 0.5 + 0.2 × 0.7 × 0.5 + 0.2 × 0.3 × 0.5
= 0.12 + 0. 07 + 0.03
= 0.22
Hence, The required probability = 0.22.
问题 25. A 和 B 轮流掷两个骰子,第一个掷出 9 的获得奖品。证明他们获胜的几率是 9 : 8。
解决方案:
According to question,
It is given that,
A and B take turns in throwing two dice.
Now, The sum of 9 can be obtained by
E = {(3, 6), (4, 5), (5, 4), (6, 3)}
P(E) = 4/36 = 1/9 and P(E‘) = 8/9
Now, P(A) = 1/9 and, P(A‘) = 8/9
P(B) = 1/9 and P(B‘) = 8/9
Now, let A starts the game
P(A wins the game)
= P(getting 9 in first throw) + P(getting 9 in third throw) + P(getting 9 in fifth throw) + ….
= 1/9 + 8/9 × 8/9 × 1/9 + 8/9 × 8/9 × 8/9 × 8/9 × 1/9 + …..
= 1/9 ×[1+ (8/9)2 + (8/9)4 + …]
= 1/9 ×[1/ (1 – (8/9)2)] [since, sum of infinite term of G.P = a/1-r]
= 9/17
P(B wins the game) = 1 – P(A wins the game) = 1 – 9/17 = 8/17
Chances of winning A:B is
= 9/17 : 8/17
= 9 : 8
Hence, chances of winning of A : B is 9 : 8.
问题 26. A、B 和 C 掷硬币。投出头的一方获胜。假设比赛可以无限期地继续下去,他们各自获胜的机会是多少?
解决方案:
According to question,
It is given that,
P(Getting head) = 1/2
P(Not getting head) = 1/2
P(A wins the game)
= P(getting head in first toss) + P(getting head in fourth toss) + P(getting head in 7th toss) + ….
= 1/2 + 1/2 × 1/2 × 1/2 × 1/2 + 1/2 × 1/2 × 1/2 × 1/2 ×1/2 ×1/2 × 1/2 + …..
= 1/2 ×[1 + (1/2)3 + (1/2)6 + …]
= 1/2 ×[1/ (1 – (1/2)3)] [since, sum of infinite term of G.P = a/1-r]
= 4/7
Now,
P(B wins the game)
= P(getting head in second toss) + P(getting head in fifth toss) + P(getting head in 8th toss) + ….
= 1/2×1/2 + 1/2 × 1/2 × 1/2 ×1/2 × 1/2 × 1/2 + 1/2 × 1/2 × 1/2 × 1/2 ×1/2 ×1/2 × 1/2 × 1/2 + …..
= 1/4 ×[1+ (1/2)3 + (1/2)6 + …]
= 1/4 × [1/ (1 – (1/2)3)] [since, sum of infinite term of G.P = a/1-r]
= 2/7
Now,
P(C wins the game) = 1 – P(A wins) – P(B wins)
= 1 – 4/7 – 2/7
= 1/7
Hence, the required probability of wining of A, B and C is 4/7, 2/7 and 1/7.
问题 27. 三个人 A、B、C 连续掷骰子,直到一个人得到“六”并赢得比赛。找出他们各自获胜的概率。
解决方案:
According to question,
It is given that,
P(Getting six) = 1/6
P(Not getting six) = 5/6
P(A wins the game)
= P(getting 6 in first throw) + P(getting 6 in fourth throw) + P(getting 6 in 7th throw) + ….
= 1/6 + 5/6 × 5/6 × 5/6 × 1/6 + 5/6 × 5/6 × 5/6 × 5/6 ×5/6 ×5/6 × 1/6 + …..
= 1/6 ×[1 + (5/6)3 + (5/6)6 + …]
= 1/6 × [1/ (1 – (5/6)3)] [since, sum of infinite term of G.P = a/1-r]
= 36/91
Now,
P(B wins the game)
= P(getting 6 in second throw) + P(getting 6 in fifth throw) + P(getting 6 in 8th throw) + ….
= 5/6 × 1/6 + 5/6 × 5/6 × 5/6 ×5/6 × 5/6 × 1/6 + 5/6 × 5/6 × 5/6 × 5/6 ×5/6 ×5/6 × 5/6 × 1/6 + …..
= 5/36 × [1 + (5/6)3 + (5/6)6 + …]
= 5/36 × [1/ (1 – (5/6)3)] [Since, sum of infinite term of G.P = a/1-r]
= 30/91
Now,
P(C wins the game) = 1 – P(A wins) – P(B wins)
= 1 – 36/91 – 30/91
= 25/91
Hence, the required probability of wining of A, B and C is 36/91, 30/91 and 25/91.
问题 28. A 和 B 轮流掷两个骰子,先掷出 10 的得奖品,说明如果 A 先掷,他们的中奖概率为 12:11。
解决方案:
According to question,
It is given that,
A and B take turns in throwing two dice.
Now, The sum of 10 can be obtained by
E = {(4, 6), (5, 5), (6, 4)}
P(E) = 3/36 = 1/12 and P(E‘) = 11/12
Now, P(A) = 1/12 and, P(A‘) = 11/12
P(B) = 1/12 and P(B‘) = 11/12
Now, let A starts the game
P(A wins the game)
= P(getting 10 in first throw) + P(getting 10 in third throw) + P(getting 10 in fifth throw) + ….
= 1/12 + 11/12 × 11/12 × 1/12 + 11/12 × 11/12 × 11/12 × 11/12 × 1/12 + …..
= 1/12 × [1 + (11/12)2 + (11/12)4 + …]
= 1/12 × [1/ (1 – (11/12)2)] [since, sum of infinite term of G.P = a/1-r]
= 12/23
P(B wins the game) = 1 – P(A wins the game) = 1 – 12/23 = 11/23
Chances of winning A:B is
= 12/23 : 11/23
= 12 : 11
Hence, chances of winning of A : B is 12 : 11.
问题 29. 袋子“A”中有 3 个红球和 5 个黑球,袋子“B”中有 2 个红球和 3 个黑球。一个球从袋子“A”中抽出,两个球从袋子“B”中抽出。找出抽出的 3 个球中 1 个是红色的,2 个是黑色的概率。
解决方案:
According to question,
It is given that,
There are 3 red and 5 black balls in the bag ‘A’ and 2 red and 3 black
balls in bag ‘B’. And, One ball is drawn from bag ‘A’ and two from bag ‘B’.
Now,
P(one red ball from bag A and 2 black ball from bag B) + P(one black ball from bag A and
one red ball from bag A and
one black ball from bag B)
= P(R1 ∩ (2B2)) + P(B1 ∩ R2 ∩ B2)
= 3/8 × 3/5 × 2/4 + 5/8 × 2/5 × 3/4 × 2
= 18/160 + 30/160 = 48/160
Required probability = 3/10.
问题 30。法蒂玛和约翰出现在同一个职位的两个空缺的面试中。 Fatima 选择的概率是 1/7,John 选择的概率是 1/5。发生的概率是多少
(i) 他们都将被选中?
(ii) 只有其中一个会被选中?
(iii) 他们都不会被选中?
解决方案:
According to Question,
It is given that,
P(F) = 1/7 and, P(F‘) = 6/7
P(J) = 1/5 and, P(J‘) = 4/5
(i) P(Both of them will be selected)
= P(F ∩ J)
= P(F) × P(J)
= 1/7 × 1/5 = 1/35
Hence, The required probability = 1/35.
(ii) P(only one of them will be selected)
= P[(F ∩ J‘) ∪ (F‘ ∩ J)]
= P(F ∩ J‘) + P(F‘ ∩ J)
= P(F) × P(J‘) + P(F‘) × P(J)
= 1/7 × 4/5 + 6/7 × 1/5
= 4/35 + 6/35 = 10/35 = 2/7
Hence, The required probability = 2/7
(iii) P(none of them will be selected)
= P(F‘∩J‘)
= P(F‘) × P(J‘)
= 6/7 × 4/5 = 24/35
Hence, The required probability = 24/35
问题 31. 一个袋子里有 8 个弹珠,其中 3 个是蓝色的,5 个是红色的。随机抽取一颗大理石,记下其颜色,然后将大理石放回袋中。再次从袋子中取出一块大理石,并记下它的颜色。找出弹珠的概率
(i) 蓝色后是红色。
(ii) 任何顺序的蓝色和红色。
(iii) 颜色相同。
解决方案:
According to question,
It is given that,
A bag contains 8 marbles of which 3 are blue and 5 are red. And, One marble is drawn at random, its colour is noted and the marble is replaced in the bag.
Now,
(i) P(Getting blue followed by red)
= P(B) × P(R)
= 3/8 × 5/8 = 15/64
Required probability = 15/64
(ii) P(Getting blue and red in any order)
= P(B) × P(R) + P(R) × P(B)
= 3/8 × 5/8 + 5/8 × 3/8
= 30/64 = 15/32
Required probability = 15/32.
(iii) P(of same color)
= P(R1) × P(R2) + P(B1) × P(B2)
= 5/8 × 5/8 + 3/8 × 3/8
= 25/64 + 9/64 = 34/64 = 17/32
Required probability = 17/32
问题 32. 一个瓮中有 7 个红球和 4 个蓝球。随机抽取两个球并替换。找出得到的概率
(i) 2个红球
(ii) 2 个蓝球
(iii) 一个红球和一个蓝球。
解决方案:
According to question,
It is given that,
An urn contains 7 red and 4 blue balls. And, Two balls are drawn at random with replacement.
Now,
(i) P(Getting 2 red balls)
= P(R1) × P(R2)
= 7/11 × 7/11 = 49/121
Required probability = 49/121
(ii) P(Getting 2 blue balls)
= P(B1) × P(B2)
= 4/11 × 4/11 = 16/121
Required probability = 16/121
(iii) P(Getting one red and one blue balls)
= P(R) × P(B) + P(B) × P(R)
= 7/11 × 4/11 + 4/11 × 7/11
= 28/121 + 28/121 = 56/121
Required probability = 56/121
问题 33. 从一副洗好的 52 张牌中抽出一张牌。结果被记录下来,卡片被替换并且牌组重新洗牌。然后从牌堆中抽出另一张牌。
(i) 两张牌同花色的概率是多少?
(ii) 第一张牌是 A 而第二张牌是红皇后的概率是多少?
解决方案:
According to question,
It is given that,
A card is drawn from a well-shuffled deck of 52 cards. The outcome is noted, the card is replaced and the deck reshuffled.
Now,
(i) We know that, There are four suit are club, spade, diamond and heart.
P(both the cards are of the same suit)
= P(Both cards are diamonds) + P(Both cards are spades) +
P(Both cards are clubs) + P(Both cards are hearts)
= 13/52 × 13/52 + 13/52 × 13/52 + 13/52 × 13/52 + 13/52 × 13/52
= 1/16 + 1/16 + 1/16 + 1/16 = 4/16 = 1/4
Required probability = 1/4
(ii) We know that, There are four ace cards and 2 red queens.
= P(Getting an ace card) × P(Getting a red queen)
= 4/52 × 2/52 = 1/338
Required probability = 1/338.
问题 34。在 100 名学生中,形成了 40 和 60 两个部分。如果你和你的朋友在 100 名学生中,以下概率是多少:(i)你们都进入同一个部分? (ii) 你们都进入不同的部分?
解决方案:
According to question,
It is given that,
Out of 100 students, two sections of 40 and 60 are formed.
Now,
(i) P(Both enter the same section)
= P(Both enter same section A) + P(Both enter the same section B)
= 40/100 × 40/100 + 60/100 × 60/100
= 4/25 + 9/25 = 13/25.
Hence, The required probability = 13/25.
(ii) P(Both enter different section)
= 1 – P(Both enter the same section)
= 1 – 13/25
= 12/25
Hence, The required probability = 12/25
问题 35. 在一场曲棍球比赛中,A 队和 B 队的进球数相同,直到比赛结束,所以为了决出胜负,裁判要求双方队长交替掷骰子,并判定该队,谁的队长获得前六名,将被宣布为获胜者。如果要求 A 队队长首发,求他们各自获胜的概率,并说明裁判的决定是否公平。
解决方案:
According to question,
It is given that,
P(Getting six) = 1/6
P(Not getting six) = 5/6
P(A wins the game)
= P(getting 6 in first throw) + P(getting 6 in third throw) + P(getting 6 in 5th throw) + ….
= 1/6 + 5/6 × 5/6 × 1/6 + 5/6 × 5/6 × 5/6 × 5/6 × 1/6 + …..
= 1/6 × [1 + (5/6)2 + (5/6)4 + …]
= 1/6 × [1/ (1 – (5/6)2)] [Since, sum of infinite term of G.P = a/1-r]
= 6/11
Now,
P(B wins the game)
= P(getting 6 in second throw) + P(getting 6 in fourth throw) + P(getting 6 in 6th throw) + ….
= 5/6 × 1/6 + 5/6 × 5/6 × 5/6 × 1/6 + 5/6 × 5/6 × 5/6 × 5/6 × 5/6 × 1/6 + …..
= 5/36 × [1 + (5/6)2 + (5/6)4 + …]
= 5/36 × [1/ (1 – (5/6)2)] [since, sum of infinite term of G.P = a/1-r]
= 5/11
Here we can see that Probabilities are not equal. So, the decision of the referee was not a fair one.