问题1.两个客户Shyam和Ekta在同一周(周二至周六)访问一家特定的商店。每个人都有可能在同一天和另一天去商店。双方都将在(i)当天访问商店的概率是多少? (ii)连续几天? (iii)不同的日子?
解决方案：
Total number of days = 5 (Tuesday, Wednesday, Thursday, Friday, Saturday)
Shyam can visit the shop in any of the 5 days.
Ekta can visit the shop in any of the 5 days.
So, total number of outcomes = (5 × 5) = 25
(i) Probability that both will visit the shop on the same day P(E);
They both will visit the shop either on Tuesday or Wednesday or Thursday or Friday or Saturday.
P(E) = 5 / 25 = 1 / 5
(ii) Probability that both will visit the shop on consecutive days P(E);
Consecutive means the following day.
So, they may visit the shop in following days; (T, W), (W, TH), (TH, F), (F, S), (W, T), (TH, W), (F, TH), (S, F)
They can visit the shop in 8 ways.
P(E) = 8 / 25
(iii) Probability that both will visit the shop on different days P(E);
We know that, P(E) + P(Ē) = 1
i.e.. Probability of visiting the shop in same day + Probability of visiting the shop in different days = 1
Probability of visiting the shop in same day P(E) = 1/ 5
Therefore, Probability that both will visit the shop on different days P(Ē) = 1 – P(E)
= 1 – (1) / 5
= 4 / 5
问题2.骰子的编号方式使其面显示数字1、2、2、3、3、6。将其掷两次，并记下两次的总得分。完成下表，该表给出了两次掷出的总得分的一些值：
|
+ |
1 |
2 |
2 |
3 |
3 |
6 |
|
1 |
2 |
3 |
3 |
4 |
4 |
7 |
|
2 |
3 |
4 |
4 |
5 |
5 |
8 |
|
2 |
|
|
|
|
5 |
|
|
3 |
|
|
|
|
|
|
|
3 |
|
|
5 |
|
|
9 |
|
6 |
7 |
8 |
8 |
9 |
9 |
12 |
总分数是几率是多少
(我什至? (ii)6? (iii)至少6个?
解决方案：
给定的表可以如下完成。
|
+ |
1 |
2 |
2 |
3 |
3 |
6 |
|
1 |
2 |
3 |
3 |
4 |
4 |
7 |
|
2 |
3 |
4 |
4 |
5 |
5 |
8 |
|
2 |
3 |
4 |
4 |
5 |
5 |
8 |
|
3 |
4 |
5 |
5 |
6 |
6 |
9 |
|
3 |
4 |
5 |
5 |
6 |
6 |
9 |
|
6 |
7 |
8 |
8 |
9 |
9 |
12 |
扔了两个骰子。因此结果总数=(6×6)= 36
(i) Probability that the total score is even P(E);
From, the above table, below are the throws in which the total score is even; (1, 1), (1, 3), (1, 3), (2, 2), (2, 2), (6, 2), (2, 2), (2, 2), (6, 2), (1, 3), (3, 3), (3, 3), (1, 3), (3, 3), (3, 3), (2, 6), (2, 6), (6, 6)
Total number of throws, in which the total score is even = 18
P(E) = 18 / 36 = 1/ 2
(ii) Probability that the total score is 6 P(E);
Events that produce total score of 6 = (3, 3), (3, 3), (3, 3), (3, 3)
Total number of throws, in which the total score is 6 = 4
P(E) = 4 / 36 = 1/ 9
(iii) Probability that the total score is atleast 6 P(E);
At least means that greater than or equal to 6 (>=).
Events that produce total score of 6 = (1, 6), (2, 6), (2, 6), (3, 3), (3, 3), (3, 6), (3, 3), (3, 3), (3, 6), (6, 1), (6, 2), (6, 2), (6, 3), (6, 3), (6, 6)
Total number of throws, in which the total score is atleast 6 = 15
P(E) = 15 / 36 = 5 / 12
问题3.一个袋子包含5个红色的球和一些蓝色的球。如果画蓝色球的概率是红色球的概率的两倍，请确定袋子中蓝色球的数量?
解决方案：
Total number of red balls = 5
Total number of blue balls = ?
Let, the total number of blue balls be = x
Then the total number of balls in the bag (both red and blue) = (5 + x)
Probability of drawing a red ball P(E) = 5 / (5 + x)
Probability of drawing a blue ball P(E) = x / (5 + x)
Given that, probability of drawing a blue ball is double that of a red ball;
(x / (5 + x)) = 2(5 / (5 + x))
=> x / (5 + x) = 10 / (5 + x)
=> 5x + x2 = 50 + 10x
=> x2 + 5x – 10x – 50 = 0
=> x2 – 5x + 50 = 0
=> (× – 10) (x + 5) = 0
Therefore,
x – 10 = 0 , x = 10
x + 5 = 0 , x = -5
As, the probability cannot be negative (probability lies betwee 0 to 1);
the x will be 10.
Total number of blue balls in the bag = 10.
问题4.一个盒子包含12个球，其中x是黑色的。如果从盒子中随机抽出一个球，那么将成为黑球的概率是多少?如果在盒子里再放6个黑球，那么现在抽黑球的概率是以前的两倍。找到x?
解决方案：
The box contains = 12 balls
Number of black balls = x
Other balls = (12 – x)
Probability that the ball drawn will be a black ball P(E);
P(E) = x / 12
After adding 6 more balls,
Now, total number of balls = 12 + 6 = 18
total black balls in the bag = (6 + x)
Probability of drawing a black ball is now double of what it was before
Probability of drawing a black ball = (6 + x) / 18
=> (6 + x) / 18 = 2 (x / 12)
=> 12 ( 6 + x ) = 18 ( 2x )
=> 2 ( 6 + x ) = 3 ( 2x )
=> 12 + 2x = 6x
=> 12 = 6x – 2x
=> 12 = 4x
=> 3 = x
Number of black balls = 3
问题5.一个罐子里装有24个弹珠，有些是绿色的，有些是蓝色的。如果从广口瓶中随机抽取大理石，则其为绿色的概率为2 /3⋅。找到广口瓶中蓝色球的数量吗?
解决方案：
Jar contains – 24 marbles
Let the number of green marbles be = x
Probability of drawing a green marble;
P(E) = 2 / 3
i.e. x / 24 = 2 / 3
=> 3x = 48
=> x = 16
Therefore, the number of blue marbles in the jar = 24 – 16 = 8