第 12 类 RD Sharma 解决方案 - 第 33 章二项分布 - 练习 33.1 |设置 2
问题 19。一个袋子包含 7 个绿色、4 个白色和 5 个红色的球。如果四个球一个接一个地抽出来,一个是红色的概率是多少?
解决方案:
Let us consider X be the number of red balls drawn from 16 balls with replacement.
So, a binomial distribution follows by X with n = 4.
Here, p = 5/16 and q = 1 – p = 11/16
P(X = r) =
P(One ball is red) = P(X = 1)
=
= 4 (5/16) (11/16)3
= (5/4) (11/16)3
问题 20。一个袋子包含 2 个白色、3 个红色和 4 个蓝色球。从袋子里随机抽取两个球。如果 X 表示抽出的两个球中有多少个白球,描述 X 的概率分布。
解决方案:
Let X denote the number of white balls when 2 balls are drawn from the bag.
So, X follows a distribution with values 0, 1, or 2.
P(X = 0) = P(All balls non – white)
=
= 42/72
= 21/36
P(X = 1) = P ( Ist ball white and IInd ball non – white)
=
= 14/36
So, P(X = 2) = P(Both balls white)
=
= 1/36
So, the tabular form is: X 0 1 2 P(X) 21/36 14/36 1/36
问题 21. 一个瓮中有四个白球和三个红球。从瓮中找出三个平局中红球数量的概率分布。
解决方案:
Given that three balls are drawn with a replacement, the number of white balls.
So, a binomial distribution follows by X with n = 3.
Here p = 3/7 and q = 4/7
P(X = r) = , r = 0, 1, 2, 3
P(X = 0) =
= 64/343
P(X = 1) =
= 144/343
P(X = 2) =
= 108/343
P(X = 3) =
= 27/343
So, the tabular form is: X 0 1 2 3 P(X) 64/343 144/343 108/343 27/343
问题 22. 求一对骰子投掷 4 次中双点数的概率分布。
解决方案:
Let us considered X denotes the number of doublets in 4 throws of a pair of dice.
So, a binomial distribution follows by X with n = 4.
Here p = No of getting (1, 1)(2, 2) . . . (6, 6)
= 6/36
= 1/6
And q = 1 – p = 5/6
P(X = r) = , r = 0, 1, 2, 3, 4
P(X = 0) =
= 625/1296
P(X = 1) =
= 500/1296
P(X = 2) =
= 150/1296
P(X = 3) =
= 20/1296
P(X = 4) =
= 1/1296
So, the distribution is: X 0 1 2 3 4 P(X) 625/1296 500/1296 150/1296 20/1296 1/1296
问题 23. 求掷骰子 3 次中 6 数的概率分布。
解决方案:
Let us considered X denotes the number of 6 in 3 tosses of a die.
So, a binomial distribution follows by X with n = 3.
Here p = 1/6, q = 1 – p = 5/6
P(X = r) = , r = 0, 1, 2, 3
So, P(X = 0) =
= 125/216
P(X = 1) =
= 75/216
P(X = 2) =
= 15/216
P(X = 3) =
= 1/216
So, the distribution is: X 0 1 2 3 P(X) 125/216 75/216 15/216 1/216
问题 24. 一枚硬币被抛 5 次。如果 X 是观察到的正面数量,求 X 的概率分布。
解决方案:
Let us considered X denotes the number of heads in 5 tosses.
So, a binomial distribution follows by X with n = 5.
Here p = 1/2 and q = 1/2
P(X = r) = , r = 0, 1, 2, 3, 4, 5
= 5Cr (2)5
P(X = 0) = 5C0 (2)5
= 1/32
P(X = 1) = 5C1 (2)5
= 5/32
P(X = 2) = 5C2 (2)5
= 10/32
P(X = 3) = 5C3 (2)5
= 10/32
P(X = 4) = 5C4 (2)5
= 5/32
P(X = 5) = 5C5 (2)5
= 1/32
So, the distribution is: X 0 1 2 3 4 5 P(X) 1/32 5/32 10/32 10/32 5/32 1/32
问题 25. 一个无偏的骰子被掷了两次。一次成功是获得一个大于 4 的数字。求成功次数的概率分布。
解决方案:
Let us consider X be the getting a number greater than 4 .
So, a binomial distribution follows by X with n = 2.
Here p = P(X > 4) = P(X = 5 or 6)
= 1/6 + 1/6
= 1/3
And q = 1 – p = 2/3
P(X = r) = , r = 0, 1, 2
So, the distribution is: X 0 1 2 P(X) 4/9 4/9 1/9
问题 26. 抛硬币时,一个人的正面赢得卢比,反面则损失一卢比。假设他投掷一次,如果他赢了就退出,但如果他在第一次投掷时输了,他会再试一次。求该人赢得卢比数的概率分布。
解决方案:
Let us consider X be the number of rupees the man wins.
So, first we assume that he gets head in the first toss.
So, the probability would be 1/2. Also, he wins Rs.1 rupee.
Now, the second possibility is that he gets a tail in the first toss.
Then he tosses again. Suppose he obtain ahead in the second toss.
Then, he wins Rs 1 rupee in the second toss but loses Rs 1 rupee in the first toss.
So, the money he won = Rs 0
So, the probability for winning Rs.0 is
= (1/2) (1/2)
= 1/4
Now, the third possibility is obtaining tail in the first toss and also tail in the second toss
Then, the money that he would win = -2 (As he loses Rs 2)
So, the probability for the third possibility = (1/2) (1/2)
= 1/4
So, the distribution is: X 0 1 2 P(X) 1/2 1/4 1/4
问题 27. 五个骰子同时掷出。如果单个骰子中出现 3、4 或 5 被认为是成功的,则求至少 3 次成功的概率。
解决方案:
Let us consider X be the occurrence of 3,4 or 5 in a single die.
So, a binomial distribution follows by X with n = 5.
Let us assume the probability of getting 3, 4 or 5 in a single die is p
Here p = 3/6 = 1/2
And q = 1 – 1/2 = 1/2
P(X = r) =
P(at least 3 successes) = P(X > 3)
= P(X = 3) + P(X = 4) + P(X = 5)
=
=
= 16/32
= 1/2
问题 28. 一家公司生产的产品包含 10% 的缺陷产品。证明在 8 个项目的样本中出现 2 个缺陷项目的概率是 .
解决方案:
Let us consider X be the number of defective items in the items produced by the company.
So, a binomial distribution follows by X with n = 8.
Here, p = 10 % = 10/100 = 1/10
And q = 1 – p = 9/10
Hence, the distribution is given by,
P(X = r) =
So, the probability of getting 2 defective items is
P(X = 2) =
=
Hence proved.
问题 29. 在普通的 52 张牌中抽取一张牌并将其替换。一张牌必须抽多少次才能使
(i) 至少有一个平均的机会画一颗心
解决方案:
Let us consider X be the probability of drawing a heart from a deck of 52 cards.
So, we get
Here, p = 13/52 = 1/4
And q = 1 – p = 1 – 1/4 = 3/4
Let us assume the card be drawn n times.
So,
P(X = r) =
Let us consider X be the number of hearts drawn from a pack of 52 cards.
So, the smallest value of n for which P(X=0) is less than 1/4
i.e., P(X = 0) < 1/4
=>
Now, put n = 1, (3/4)1 not less than 1/4
n = 2, (3/4)2 not less than 1/4
n = 3, (3/4)3 not less than 1/4
So, smallest value of n = 3.
(ii) 画心的概率大于 3/4
解决方案:
Given that the probability of drawing a heart > 3/4.
so, 1 – P(X = 0) > 3/4
For n = 1, (3/4)1 not less than 1/4.
n = 2, (3/4)2 not less than 1/4
n = 3, (3/4)3 not less than 1/4
n = 4, (3/4)4 not less than 1/4
n = 5, (3/4)5 not less than 1/4
So, card must be drawn 5 times.
问题 30. 数学系有 8 名研究生助理,他们被分配到同一个办公室。每个助理在家学习的可能性与在办公室学习的可能性一样。办公室必须有多少张桌子才能让每个助理至少 90% 的时间都有一张桌子?
解决方案:
Let us consider k denotes the number of desks and X denotes the number of graduate assistants in the office.
So, a binomial distribution follows by X with n = 8.
Here p = 1/2 and q = 1/2.
So,
=> P(X < k) > 90%
=> P\left( X < k \right) > 0.90
=> P\left( X > k \right) < 0.10
=> P(X = k + 1, k + 2, . . . . 8) < 0 . 10
Therefore, P(X > 6) = P(X = 7 or X = 8)
Now, P(X > 5) = P(X = 6, X = 7 or X = 8) = 0.15
P(X > 6) < 0.10
Hence, if there are 6 desks then there is at least 90% chance for every graduate to get a desk.
问题 31. 一枚无偏的硬币被抛 8 次。通过使用二项分布,找出至少出现 6 个正面的概率。
解决方案:
Let us considered X be the number of heads in tossing the coin 8 times.
So, a binomial distribution follows by X with n = 8.
Here p = 1/2 and q = 1/2
Hence,
P(X = r) = , r = 0, 1, 2, 3, 4, 5, 6, 7, 8
So, the required probability is
P(X > 6) = P(X = 6) + P(X = 7) + P(X = 8)
=
=
= 37/256
问题 32. 同时抛 6 个硬币。求得到的概率:
(一) 3个头
解决方案:
Let us considered X be the number of heads obtained in tossing 6 coins.
So, a binomial distribution follows by X with n = 6.
Here p = 1/2 and q = 1/2
Hence,
P(X = r) = , r = 0, 1, 2, 3, 4, 5, 6
=
P(getting 3 heads) = P(X = 3)
=
= 20/64
= 5/16
(ii) 无头
解决方案:
P(getting no head) = P(X = 0)
=
= (1/2)6
= 1/64
(iii) 至少一名
解决方案:
P(getting at least 1 head) = P(X > 1)
= 1 – P(X = 0)
= 1 – 1/64
= 63/64
问题 33。假设插入某种类型的设备中的无线电管工作超过 500 小时的概率为 0.2。如果我们随机测试 4 个管子,其中恰好三个管子函数超过 500 小时的概率是多少?
解决方案:
Let us considered X be the number of tubes that function for more than 500 hours.
So, a binomial distribution follows by X with n = 4.
Let us considered p be the probability that the tubes function more than 500 hours.
Here , p = 0.2, q = 0.8
Hence,
P(X = r) = 4Cr (0.2)r (0.8)4-r, r = 0, 1, 2, 3, 4
So, the required probability is
P(X = 3) = 4 (0.2)3 (0.8)
= 0.0256
问题 34. 某种部件在给定的冲击测试中幸存下来的概率是 3/4。
(i) 找出在 5 个被测试的组件中恰好有 2 个能够存活的概率。
解决方案:
Let us considered X be the number of components that survive shock.
So, a binomial distribution follows by X with n = 5.
Let us considered p be the probability that a certain kind of
component will survive a given shock test.
So, p = 3/4 and q = 1/4
Hence,
P(X = r) = , r = 0, 1, 2, 3, 4, 5
P(exactly 2 will survive} ) = P(X = 2)
=
=
= 0.0879
(ii) 找出在 5 个被测试的组件中最多有 3 个能够存活的概率。
解决方案:
Let us considered X be the number of components that survive shock.
So, a binomial distribution follows by X with n = 5.
Let us considered p be the probability that a certain kind of
component will survive a given shock test.
So, p = 3/4 and q = 1/4
Hence,
P(X = r) = , r = 0, 1, 2, 3, 4, 5
P(at most 3 will survive) = P(X < 3)
= P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
=
=
=
= 376/1024
= 0.3672
问题 35。假设从飞机上投下的炸弹击中某个目标的概率为 0.2。如果投下 6 颗炸弹,求概率:
(i) 正好 2 击中目标。
解决方案:
Let us considered X be the number of bombs that hit the target and
p be the probability that a bomb dropped from an aeroplane will strike the target.
So, a binomial distribution follows by X with n = 6
p = 0.2 and q = 0.8
Hence,
P(X = r) =
P(exactly 2 will strike the target) = P(X = 2)
=
= 0.2458
(ii) 至少有 2 人会击中目标
解决方案:
Let us considered X be the number of bombs that hit the target and
p be the probability that a bomb dropped from an aeroplane will strike the target.
So, a binomial distribution follows by X with n = 6.
p = 0.2 and q = 0.8
Hence,
P(X = r) =
P(at least 2 will strike the target) = P(X > 2)
= 1 – [P(X = 0) + P(X = 1)]
= 1 – (0.8)6 – 6 (0.2) (0.8)5
= 1 – 0.2621 – 0.3932
= 0 . 3447
问题 36. 众所周知,60% 的小鼠接种了血清后可免受某种疾病的侵害。如果接种了 5 只小鼠,求以下概率:
(i) 没有人感染这种疾病。
解决方案:
Let us considered X be the number of mice that contract the disease and
p be the probability of mice that contract the disease.
So, a binomial distribution follows by X with n = 5.
p = 0.4 and q = 0.6
Hence,
P(X = r) = , r = 0, 1, 2, 3, 4, 5
P(X = 0) =
= (0.6)5
= 0.0778
(ii) 3 人以上感染该疾病。
解决方案:
Let us considered X be the number of mice that contract the disease and
p be the probability of mice that contract the disease.
So, a binomial distribution follows by X with n = 5.
p = 0.4 and q = 0.6
Hence,
P(X = r) = , r = 0, 1, 2, 3, 4, 5
P(X > 3) = P(X = 4) + P(X = 5)
=
= 0.0768 + 0.01024
= 0.08704