Class 10 NCERT解决方案-第15章概率–练习15.1 |套装2

问题13：一次掷一次骰子。找到获得的可能性：

(i)质数

(ii)一个介于2到6之间的数字

(iii)一个奇数。

解决方案：

Here, total number of possible outcomes = 6

(i) P(E) = Probability of getting a prime number.

Prime numbers = {2,3,5}

P(E) = $\frac{3}{6} = \frac{1}{2}$ ………………. (From Theorem 1)

(ii) P(E) = Probability of getting a number between 2 and 6.

Numbers between 2 and 6 = {3,4,5}

P(E) = $\frac{3}{6} = \frac{1}{2}$ ………………. (From Theorem 1)

(iii) P(E) = Probability of getting an odd number.

Odd numbers = {1,3,5}

P(E) = $\frac{3}{6} = \frac{1}{2}$ ………………. (From Theorem 1)

问题14：一张纸牌是从一张52张纸牌中抽调得很好的。找到获得的可能性

(i)红色之王

(ii)面卡

(iii)红脸卡

(iv)心中的杰克

(v)一把铁锹

(vi)钻石皇后

解决方案：

Here, total number of possible outcomes = 52

(i) P(E) = Probability of getting a king of red colour.

Number king of red colour = 2

P(E) = $\frac{2}{52} = \frac{1}{26}$ ………………. (From Theorem 1)

(ii) P(E) = Probability of getting a face card.

Number face cards = 12

P(E) = $\frac{12}{52} = \frac{3}{13}$ ………………. (From Theorem 1)

(iii) P(E) = Probability of getting a red face card.

Number red face cards = 6

P(E) = $\frac{6}{52} = \frac{3}{26}$ ………………. (From Theorem 1)

(iv) P(E) = Probability of getting the jack of hearts.

Number of jack of hearts = 1

P(E) = $\frac{1}{52}$ ………………. (From Theorem 1)

(v) P(E) = Probability of getting a queen of diamond.

Number of queen of diamonds = 1

P(E) = $\frac{1}{52}$ ………………. (From Theorem 1)

问题15：五张牌(钻石的十张，千斤顶，王后，国王和王牌)洗净后的脸朝下。然后随机抽取一张卡。

(i)这张卡是皇后的概率是多少?

(ii)如果女王被抽走并放在一边，那么第二张牌被拾取的概率是多少?

(a)王牌?

(b)女王?

解决方案：

Here, total numbers of cards = 5

(i) P(E) = Probability of getting a queen.

P(E) = $\frac{1}{5}$ ………………. (From Theorem 1)

(ii) When the queen is put aside, then Here, total numbers of remaining cards = 4

(a) P(E) = Probability of getting an ace.

P(E) = $\frac{1}{4}$………………. (From Theorem 1)

(b) P(E) = Probability of getting a queen.

P(E) =$\frac{0}{4}$ = 0 ………………. (From Theorem 1)

问题16。12支有缺陷的笔意外地与132支好笔混合在一起。不能只看笔并判断它是否有缺陷。从这批中随机抽出一支笔。确定取出笔是好的笔的可能性。

解决方案：

Number of deflective pens = 12

Number of good pens = 132

Total number of pens = 12+132 = 144

Hence, total number of possible outcomes = 144

P(E) = Probability of getting a good pen.

P(E) = $\frac{132}{144} = \frac{11}{12}$ ………………. (From Theorem 1)

问题17。(i)很多20个灯泡中包含4个有缺陷的灯泡。从抽签中随机抽取一个灯泡。该灯泡有故障的可能性是多少?

(ii)假设未损坏的灯泡已被拉入且未更换。现在从其余部分中随机抽出一个灯泡。该灯泡无故障的概率是多少?

解决方案：

(i) Numbers of defective bulbs = 4

The total numbers of bulbs = 20

Hence, total number of possible outcomes = 20

P(E) = Probability of getting a defective bulb

P(E) = $\frac{4}{20} = \frac{1}{5}$ ………………. (From Theorem 1)

(ii) As a non-defective bulb is drawn, then the total numbers of bulbs left are 19

Hence, the total numbers of outcomes = 19

Numbers of non-defective bulbs = 19-4 = 15

P(E) = Probability of getting a non-defective bulb

P(E) = $\frac{15}{19}$ ………………. (From Theorem 1)

问题18.一个盒子包含90张从1到90的光盘。如果从盒子中随机抽出一张光盘，请找出它承受的概率。

(i)一个两位数的数字

(ii)一个完美的平方数

(iii)一个可被5整除的数字。

解决方案：

The total numbers of discs (outcomes) = 90

(i) Total number of discs having two-digit numbers = 81

(As 1 to 9 are single digit numbers)

P(E) = Probability of getting a two digit-number

P(E) = $\frac{81}{90} = \frac{9}{10}$ ………………. (From Theorem 1)

(ii) Total numbers of perfect square number = 9

{1, 4, 9, 16, 25, 36, 49, 64 and 81}

P(E) = Probability of getting a perfect square

P(E) = $\frac{9}{90} = \frac{1}{10}$ ………………. (From Theorem 1)

(iii) Total numbers which are divisible by 5 = 18

{5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85 and 90}

P(E) = Probability of getting a number divisible by 5.

P(E) = $\frac{18}{90} = \frac{1}{5}$ ………………. (From Theorem 1)

问题19.一个孩子有一个死者，其六个面孔显示以下字母：

1

2

3

4

5

6

掷一次骰子。得到的几率是多少

(i)A?

(ii)D?

解决方案：

The total numbers of outcomes = 6

(i) P(E) = Probability of getting A

Total number of ‘A’ in dice = 2

P(E) = $\frac{2}{6} = \frac{1}{3}$ ………………. (From Theorem 1)

(ii) P(E) = Probability of getting D

Total number of ‘D’ in dice = 1

P(E) = $\frac{1}{6}$ ………………. (From Theorem 1)

问题20.假设您将骰子随机放置在图15.6所示的矩形区域上。它将落入直径为1m的圆内的概率是多少?

解决方案：

Here, the area of the rectangle is the possible outcome and,

The area of the circle will be the favourable outcome.

So, the area of the rectangle = (length × breadth) = (3×2) = 6 m²

and, the area of the circle = πr² = π(½)² = π/4 m²

P(E) = The probability that die will land inside the circle

P(E) = $\frac{(\frac{π}{4})}{6} = \frac{π}{24}$

问题21：很多笔由144个圆珠笔组成，其中有20个有缺陷，而其他则为好。 Nuri会买一支好笔，但如果有缺陷则不会买。店主随机抽一支笔给她。概率是多少

(i)她会买吗?

(ii)她不会购买吗?

解决方案：

Number of defective pens = 20

Total number of pens (outcomes)= 144

(i) P(E) = The probability that she will buy =  The probability of getting a good pens

Number of good pens = 144-20 = 124

P(E) = $\frac{124}{144} = \frac{31}{36}$ ………………. (From Theorem 1)

(ii) P(E) = The probability that she will not buy =  The probability of getting a defective pens

P(E) = $\frac{20}{144} = \frac{5}{36}$ ………………. (From Theorem 1)

问题22。请参阅示例13。

(i)填写下表：

(ii)一个学生争辩说，“有11种可能的结果2、3、4、5、6、7、8、9、10、11和12。因此，每个结果的概率为1/11。您是否同意这种说法?证明你的答案。

解决方案：

When two dices are thrown, then the total outcomes = 36

(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)

(i) P(E) = Probability of having sum as 2 = $\frac{1}{36}$ (Given)

Sample space is (1,1)

P(E) = Probability of having sum as 3

Sample space are (1,2) and (2,1)

P(E) = $\frac{2}{36} = \frac{1}{18}$

P(E) = Probability of having sum as 4

Sample space are (1,3), (3,1), and (2,2)

P(E) = $\frac{3}{36} = \frac{1}{12}$

P(E) = Probability of having sum as 5

Sample space are (1,4), (4,1), (2,3), and (3,2)

P(E) = $\frac{4}{36} = \frac{1}{9}$

P(E) = Probability of having sum as 6

Sample space are (1,5), (5,1), (2,4), (4,2), and (3,3)

P(E) = $\frac{5}{36}$

P(E) = Probability of having sum as 7

Sample space are (1,6), (6,1), (5,2), (2,5), (4,3), and (3,4)

P(E) = $\frac{6}{36} = \frac{1}{6}$

P(E) = Probability of having sum as 8

Sample space are (2,6), (6,2), (3,5), (5,3), and (4,4)

P(E) = $\frac{5}{36}$

P(E) = Probability of having sum as 9

Sample space are (3,6), (6,3), (4,5), and (5,4)

P(E) = $\frac{4}{36} = \frac{1}{9}$

P(E) = Probability of having sum as 10

Sample space are (4,6), (6,4), and (5,5)

P(E) = $\frac{3}{36} = \frac{1}{12}$

P(E) = Probability of having sum as 11

Sample space are (5,6), and (6,5)

P(E) = $\frac{2}{36} = \frac{1}{18}$

P(E) = Probability of having sum as 12

Sample space is (6,6)

P(E) = $\frac{1}{36}$

(ii) The argument is not correct as it contradicts the verified solution in (i) that the number of all possible outcomes is 36 and not 11.

问题23：一场游戏包括将1卢比硬币扔3次，并每次记录其结果。如果所有掷骰子都产生相同的结果(即三头或三尾)，则哈尼夫将获胜，否则将输掉。计算Hanif输掉比赛的概率。

解决方案：

The total number of outcomes = 8, as follows

HHH

HHT

HTT

HTH

THT

THH

TTH

TTT

Total outcomes in which Hanif will lose the game = 6 

HHT

HTT

HTH

THT

THH

TTH

P(E) = The probability that he will loose

P(E) = $\frac{6}{8} = \frac{3}{4}$

问题24.掷两次骰子。概率是多少

(i)5次都不会出现吗?

(ii)5个至少会出现一次?

[提示：两次掷骰子和同时掷两个骰子被视为同一实验]

解决方案：

When two dices are thrown, then the total outcomes = 36

(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)

(i) P(E) = Probability that 5 will not come up either time

5 will not come up either time = (36-11) = 25

P(E) = $\frac{25}{36}$ ………………..(From Theorem 1)

(ii) P(E) = Probability that 5 will not come up at least once

P(E) = $\frac{11}{36}$ …………………(From Theorem 1)

问题25.以下哪个参数正确，哪些不正确?解释你的回答。

(i)如果同时投掷两枚硬币，则有三种可能的结果-两只头，两只尾巴或每只一个。因此，对于这些结果中的每一个，概率都是1/3。

(ii)如果掷出骰子，则有两种可能的结果-奇数或偶数。因此，获得奇数的概率为1/2。

解决方案：

(i) The total number of outcomes = 4, as follows

HH

HT

TH

TT

So, P (getting two heads) = 1/4

P (getting two tails) = 1/4

and, P (getting one of the each) = $\frac{2}{4} = \frac{1}{2}$

Hence, this statement is INCORRECT.

(ii) Total outcomes = 6

1

2

3

4

5

6

P(Odd) = Probability of getting an odd number.

Odd numbers = {1,3,5}

P(Odd) = $\frac{3}{6} = \frac{1}{2}$ ………………(I). (From Theorem 1)

From (I) we conclude that,

The probability of getting an odd number is 1/2.

Hence, this statement is CORRECT.