第 12 类 RD Sharma 解决方案 - 第 33 章二项分布 - 练习 33.1 |设置 1

问题 1. 大量商品中有 6% 的次品。求 8 个项目的样本中包含不超过一个缺陷项目的概率。

解决方案：

Let us consider X be the number of defective items in a sample of 8 items. 

So, a binomial distribution follows by x with n = 8.

and Probability of getting a defective item(p) = 0.06 

The probability of getting a defective item (1 – p) = 0.94

So, P(X = r) = ⁸C_r (0.06)^r (0 . 94 )^{8 – r}, where r = 0, 1, 2, 3, . . . 8

So, the required probability is

= P (X < 1) 

= P(X = 0) + P(X = 1)

= ⁸C₀ (0.06)⁰ (0.94)^8-0 + ⁸C₁ (0.06)¹ (0.94 )^8-1

= (0.94)⁸ + 8 (0.06) (0.94)⁷ 

= (0.94)⁷ {0.94 + 0.48}

= 1.42 (0.94)⁷

问题 2. 一枚硬币被抛 5 次。获得至少 3 个正面的概率是多少？

解决方案：

Let us consider X be the number of heads in 5 tosses.

So, a binomial distribution follows by x with n = 5

Probability of getting a head(p) = 1/2

And q = 1 – p = 1 – 1/2 = 1/2

P(X = r) = ⁵C_r (1/2)^r (1/2)^5-r , where r = 0, 1, 2 . . . 5

So, the required probability is

= P(X > 3)

= P(X = 3) + P(X = 4) + P(X = 5)

= ⁵C₃ (1/2)³ (1/2)^5-3 + ⁵C₄ (1/2)¹ (1/2)^5-1 + ⁵C₅ (1/2)⁵ (1/2)^5-0

= 10 (1/2)⁵ + 5 (1/2)⁵ + 1 (1/2)⁵

= (1/2)⁵ (10 + 5 + 1)

= (1/2)⁵ (16)

= 1/2

问题 3. 一枚硬币被抛 5 次。尾巴出现奇数次的概率是多少？

解决方案：

Let us consider X be the number of tails when a coin is tossed 5 times.

So, a binomial distribution follows by x with n  = 5

Probability of getting head(p) = 1/2.

Also, q = 1 – p = 1/2

P(X = r) = ⁵C₃ (1/2)^r (1/2)^n-r = ⁵C_r (1/2)⁵

So, the required probability is

= P(X = 1) + P(X = 3) + P(X = 5)

= ⁵C₁ (1/2)⁵ + ⁵C₃ (1/2)⁵ + ⁵C₅ (1/2)⁵

= (1/2)⁵ [5 + 10 + 1]

= 16/32

= 1/2

问题 4. 一对骰子被掷了 6 次。如果总共获得 9 次被认为是成功的，那么至少 5 次成功的概率是多少？

解决方案：

Let us consider X be the number of successes in 6 throws of the two dice.

So the probability of success is equal to the probability of getting a total of 9  

= Probability of getting (3, 6), (4, 5), (5, 4), (6, 3) from 36 outcomes

Here, p = 4/36 = 1/9

Also q = 1 – p = 8/9 and n = 6

Now, a binomial distribution X follows with n = 6, p = 1/9, and q = 8/9

P(X = r) = ⁶C_r (1/9)^r (8/9)^6-r

The required probability is

= P(X > 5)

= P(X = 5) + P(X = 6)

= $^{6}{}{C}_5 \left( \frac{1}{9} \right)^5 \left( \frac{8}{9} \right)^{6 - 5} + ^{6}{}{C}_6 \left( \frac{1}{9} \right)^6 \left( \frac{8}{9} \right)^{6 - 6}$

= $\frac{6(8) + 1}{9^6}$

= 49/9⁶

问题 5. 一枚公平的硬币被抛 8 次，求概率：

(i) 正好 5 个正面。

解决方案：

Let us consider X be the number of heads obtained when a fair is tossed 8 times.

Now, a binomial distribution X follows with n  = 8.

Here, p = 1/2 and q = 1 – 1/2 = 1/2

So, P(X = r) = $^8 C_r \left( \frac{1}{2} \right)^{8 - r} \left( \frac{1}{2} \right)^r$

= ⁸C_r (1/2)⁸ , where r = 0, 1, 2, . . . , 8

So, the probability of obtaining exactly 5 heads is 

P(X = 5) = ⁸C₅ (1/2)⁸

= 56/256

= 7/32

(ii) 至少六个头。

解决方案：

Let us consider X be the number of heads obtained when a fair is tossed 8 times.

Now, a binomial distribution X follows with n = 8.

Here, p = 1/2 and q = 1 – 1/2 = 1/2

So, P(X = r) = $^8 C_r \left( \frac{1}{2} \right)^{8 - r} \left( \frac{1}{2} \right)^r$

= ⁸C_r (1/2)⁸ , where r = 0, 1, 2, . . . , 8

So, the probability of obtaining exactly 5 heads is 

P(X = 5) = ⁸C₅ (1/2)⁸

= 56/256

= 7/32

(iii) 最多六个头。

解决方案：

Let us consider X denotes the number of heads obtained when a fair is tossed 8 times.

Now, a binomial distribution X follows with n = 8.

Here p = 1/2 and q = 1 – 1/2 = 1/2

P(X = r) = $^8 C_r \left( \frac{1}{2} \right)^{8 - r} \left( \frac{1}{2} \right)^r$

= ⁸C_r (1/2)⁸ , r = 0, 1, 2, . . . , 8

so the probability of obtaining at most 6 heads is

P (X < 6) = 1 – [P(X = 7) + P(X = 8)]

= $1 - \left[ {}^8 C_7 \left( \frac{1}{2} \right)^8 +^8 C_8 \left( \frac{1}{2} \right)^8 \right]$

= 1 – (8/256 + 1/256)

= 1 – 9/256

= 247/256

问题 6. 找出 4 在两次投掷公平骰子中至少出现一次的概率。

解决方案：

Let us consider X denotes the probability of getting 4 in two tosses of a fair die.

Now, a binomial distribution X follows with n = 2.

Here, p = 1/6 and q = 5/6

P(X = r) = $^{2}{}{C}_r \left( \frac{1}{6} \right)^r \left( \frac{5}{6} \right)^{2 - r}$

So, the probability of obtaining 4 at least once is

P(X > 1) = 1 – P(X = 0)

= 1 – $^{2}{}{C}_0 \left( \frac{1}{6} \right)^0 \left( \frac{5}{6} \right)^{2 - 0}$

= 1 – 25/36

= 11/36

问题 7. 一枚硬币被抛 5 次。头部出现偶数次的概率是多少？

解决方案：

Let us consider X denotes the number of heads that appear when a coin is tossed 5 times.

Now, a binomial distribution X follows with n = 5 

Here p = q = 1/2.

P(X = r) = $^{5}{}{C}_r \left( \frac{1}{2} \right)^r \left( \frac{1}{2} \right)^{5 - r}$

= ⁵C_r (1/2)⁵

P (head appears an even number of  times) = P(X = 0) + P(X = 2) + P(X = 4)

= $^{5}{}{C}_0 \left( \frac{1}{2} \right)^5 + ^{5}{}{C}_2 \left( \frac{1}{2} \right)^5 + ^{5}{}{C}_4 \left( \frac{1}{2} \right)^5$

= $\frac{1 + 10 + 5}{2^5}$

= 16/32

= 1/2

问题 8. 一个人击中目标的概率是 1/4。如果他开火 7 次，他至少击中目标两次的概率是多少？

解决方案：

Let us consider X denotes the number of times the target is hit. 

Now, a binomial distribution X follows with n = 7.

Here, p = 1/4 and q = 3/4.

P(X = r) = $^{7}{C}_r \left( \frac{1}{4} \right)^r \left( \frac{3}{4} \right)^{7 - r}$

P(hit the target at least twice) = P(X > 2) 

= 1 – {P(X = 0) + P(X = 1)}

= $1 -^{7}{}{C}_0 \left( \frac{1}{4} \right)^0 \left( \frac{3}{4} \right)^{7 - 0} - ^{7}{}{C}_1 \left( \frac{1}{4} \right)^1 \left( \frac{3}{4} \right)^{7 - 1}$

= 1 – (3/4)⁷ – 7 (1/4) (3/4)⁶

= 1 – 1/16384(2187 + 5103)

= 1 – 3645/8192

= 4547/8192

问题 9。假设平日下午 2 点到 3 点之间拨打的 15 个电话号码中，平均有一个电话号码占线。如果拨打 6 个随机选择的电话号码，其中至少 3 个会占线的概率是多少？

解决方案：

Let us consider X denotes the number of busy calls for 6 randomly selected telephone numbers.

Now, a binomial distribution X follows with n = 6.

Here, p = one out of 15 = 1/15

And q = 1 – 1/15 = 14/15

P(X = r) = $^{6}{}{C}_r \left( \frac{1}{15} \right)^r \left( \frac{14}{15} \right)^{6 - r}$

So, the probability that at least 3 of them are busy is 

P(X > 3) = 1 – {P(X = 0) + P(X = 1) + P(X = 2)}

= $1 - \left\{ ^{6}{}{C}_0 \left( \frac{1}{15} \right)^0 \left( \frac{14}{15} \right)^{6 - 0} + ^{6}{}{C}_1 \left( \frac{1}{15} \right)^1 \left( \frac{14}{15} \right)^{6 - 1} + ^{6}{}{C}_2 \left( \frac{1}{15} \right)^2 \left( \frac{14}{15} \right)^{6 - 2} \right\}$

= $1 - \left\{ \left( \frac{14}{15} \right)^6 + \frac{6}{15} \left( \frac{14}{15} \right)^5 + \frac{1}{15} \left( \frac{14}{15} \right)^4 \right\}$

问题 10. 如果在一个无偏的骰子中掷出 5 或 6 个是成功的，并且随机变量 X 表示掷出 6 次骰子的成功次数，求 P (X ≥ 4)。

解决方案：

Let us consider X be the number of successes. That is getting 5 or 6 in a throw of die in 6 throws.

Now, a binomial distribution X follows with n = 6

Here, p = of getting 5 or 6 = 1/6 + 1/6 = 1/3

 And q = 1 – p = 2/3

P(X = r) = $^{6}{}{C}_r \left( \frac{1}{3} \right)^r \left( \frac{2}{3} \right)^{6 - r}$

P(X > 4) = P(X = 4) + P(X = 5) + P(X = 6)

= $^{6}{}{C}_4 \left( \frac{1}{3} \right)^4 \left( \frac{2}{3} \right)^{6 - 4} +^{6}{}{C}_5 \left( \frac{1}{3} \right)^5 \left( \frac{2}{3} \right)^{6 - 5} + ^{6}{}{C}_6 \left( \frac{1}{3} \right)^6 \left( \frac{2}{3} \right)^{6 - 6}$

= (1/3⁶) (60 + 12 + 1)

= 73/729

问题 11. 八枚硬币同时投掷。找到获得至少六个正面的机会。

解决方案：

Let us consider X denotes the number of heads in tossing 8 coins.

Now, a binomial distribution X follows with n = 8.

Here, p = 1/2 and q = 1/2

P(X = r) = $^{8}{}{C}_r \left( \frac{1}{2} \right)^r \left( \frac{1}{2} \right)^{8 - r}$

= ⁸C_r (1/2)⁸

So, the probability of getting at least 6 heads is 

P(X > 6) = P(X = 6) + P(X = 7) + P(X = 8)

= $^{8}{}{C}_6 \left( \frac{1}{2} \right)^8 + ^{8}{}{C}_7 \left( \frac{1}{2} \right)^8 + ^{8}{}{C}_8 \left( \frac{1}{2} \right)^8$

= 1/2⁸(28 + 8 + 1)

= 37/256

问题 12. 从一副洗好的 52 张牌中依次抽出 5 张牌并替换。

(i) 五张牌都是黑桃的概率是多少？

解决方案：

Let us consider X be the number of spade cards when 5 cards are drawn with replacement.

Now, a binomial distribution X follows with n = 5.

p = 13/52 = 1/4 and q = 1 – p = 3/4

P(X = r) = $^{5}{}{C}_r \left( \frac{1}{4} \right)^r \left( \frac{3}{4} \right)^{5 - r}$

P(All cards are spades) = P(X = 5)

= $^{5}{}{C}_5 \left( \frac{1}{4} \right)^5 \left( \frac{3}{4} \right)^0$

= 1/1024

(ii) 只有 3 张牌是黑桃的概率是多少？

解决方案：

Let us consider X be the number of spade cards when 5 cards are drawn with replacement.

Now, a binomial distribution X follows with n = 5. 

Here, p = 13/52 = 1/4 and q = 1 – p = 3/4

P(X = r) = $^{5}{}{C}_r \left( \frac{1}{4} \right)^r \left( \frac{3}{4} \right)^{5 - r}$

P(only 3 cards are spades) = P(X = 3)

= $^{5}{}{C}_3 \left( \frac{1}{4} \right)^3 \left( \frac{3}{4} \right)^2$

= (1/1024) (90)

= 45/512

(iii) 没有黑桃的概率是多少？

解决方案：

Let us consider X be the number of spade cards when 5 cards are drawn with replacement.

Now, a binomial distribution X follows with n = 5.

Here p = 13/52 = 1/4 and q = 1 – p = 3/4.

P(X = r) = $^{5}{}{C}_r \left( \frac{1}{4} \right)^r \left( \frac{3}{4} \right)^{5 - r}$

P(none is a spade) = P(X = 0)

= $^{5}{}{C}_0 \left( \frac{1}{4} \right)^0 \left( \frac{3}{4} \right)^5$

= 243/1024

问题 13. 一个袋子包含 7 个红球、5 个白球和 8 个黑球。四个球一一抽出，有替换。

(i) 没有人是白人的概率是多少？

解决方案：

Let us consider X denotes the number of white balls drawn when 4 balls are drawn with replacement.  

Now, a binomial distribution X follows with n = 4.

Here, the probability for a white ball(p) = 5/20 = 1/4

And q = 1 – p = 3/4

P(X = r) = $^{4}{}{C}_r \left( \frac{1}{4} \right)^r \left( \frac{3}{4} \right)^{4 - r}$

So, the probability that none is white is

P(X = 0) = $^{4}{}{C}_0 \left( \frac{1}{4} \right)^0 \left( \frac{3}{4} \right)^{4 - 0}$

= 81/256

(ii) 没有人是白人的概率是多少？

解决方案：

Let us consider X denotes the number of white balls drawn when 4 balls are drawn with replacement.  

Now, a binomial distribution X follows with n = 4.

Here, the probability for a white ball(p) = 5/20 = 1/4

And q = 1 – p = 3/4

P(X = r) = $^{4}{}{C}_r \left( \frac{1}{4} \right)^r \left( \frac{3}{4} \right)^{4 - r}$

so, the probability that none is white is 

P(X = 0) = $^{4}{}{C}_0 \left( \frac{1}{4} \right)^0 \left( \frac{3}{4} \right)^{4 - 0}$

= 81/256

(iii) 任何两个都是白色的概率是多少？

解决方案：

Let us consider X denotes the number of white balls drawn when 4 balls are drawn with replacement.  

Now, a binomial distribution X follows with n = 4.

So, the probability for a white ball(p) = 5/20 = 1/4

And q = 1 – p = 3/4

P(X = r) = $^{4}{}{C}_r \left( \frac{1}{4} \right)^r \left( \frac{3}{4} \right)^{4 - r}$

So, the probability that any two are white is 

P(X = 2) = $^{4}{}{C}_2 \left( \frac{1}{4} \right)^2 \left( \frac{3}{4} \right)^{4 - 2}$

= 54/256

= 27/128

问题 14. 一个盒子里有 100 张票，每张票上的数字是 1 到 100 中的一个。如果从盒子里连续抽出 5 张票并放回，求所有票上的数字能被 10 整除的概率。

解决方案：

Let us consider X denotes the variable representing number on the ticket bearing a number divisible by 10 out of the 5 tickets drawn.

Now, a binomial distribution X follows with n = 5.

Here, the probability of getting a ticket bearing number divisible by 10(p) = 10/100

= 1/10

And q = 1 – p = 9/10

P(X = r) = $^{5}{}{C}_r \left( \frac{1}{10} \right)^r \left( \frac{9}{10} \right)^{5 - r}$

So, the probability that all the tickets bear numbers divisible by 10 is

P(X = 5) = $^{5}{}{C}_5 \left( \frac{1}{10} \right)^5 \left( \frac{9}{10} \right)^{5 - 5}$

= (1/10)⁵ (9/10)⁰

= (1/10)⁵

问题 15. 一个袋子里有 10 个球，每个球都标有从 0 到 9 的一个数字。如果从袋子里连续抽出四个球并放回，那么一个球没有标数字 0 的概率是多少？

解决方案：

Let us consider X denotes the number of balls marked with the digit 0 when 4 balls are drawn successfully with replacement.

Now, a binomial distribution X follows with n = 4.

So, the probability that a ball randomly drawn bears digit 0(p) = 1/10

And q = 1 – p = 9/10

P(X = r) = $^{4}{}{C}_r \left( \frac{1}{10} \right)^r \left( \frac{9}{10} \right)^{4 - r}$

P(none bears the digit 0) = P(X = 0)

= $^{4}{}{C}_0 \left( \frac{1}{10} \right)^0 \left( \frac{9}{10} \right)^{4 - 0}$

= (9/10)⁴

问题 16。在大量物品中，有 5% 的物品有缺陷。 10 个项目的样本中包含不超过一个缺陷项目的概率是多少？

解决方案：

Let us consider X be the number of defective items in a sample of 10 items.

Now, a binomial distribution X follows with n = 10.

And the probability of defective items(p) = 5% = 0.05 

And q = 1 – p = 0.95

P(X = r) = ¹⁰C_r (0. 05)^r (0.95)^10-r

So, the probability(sample of 10 items will include not more than one defective item) is 

P(X < 1) = P(X = 0) + P(X = 1) 

= 10C0 (0.05)⁰ (0.95 )^{0-0 + ¹⁰C₁ (0.05)¹ (0.95)^10-1

= (0.95)⁹ (0.95 + 0.5)

= 1.45 (0.95)⁹

问题 17. 工厂生产的灯泡使用 150 天后熔断的概率为 0.05。

(i) 求 5 个这样的灯泡中没有一个在使用 150 天后熔断的概率。

解决方案：

Let us consider X be the number of bulbs that fuse after 150 days.

Now, a binomial distribution X follows with n = 5.

Here, p = 0.05 and  q = 0.95

Or p = 1/20 and q = 19/20

P(X = r) = $^{5}{}{C}_r \left( \frac{1}{20} \right)^r \left( \frac{19}{20} \right)^{5 - r}$

So, the probability (none will fuse after 150 days of use) is 

P(X = 0) = $^{5}{}{C}_0 \left( \frac{1}{20} \right)^0 \left( \frac{19}{20} \right)^{5 - 0}$

= (19/20)⁵

(ii) 找出在 5 个这样的灯泡中，使用 150 天后不超过一个会熔断的概率。

解决方案：

Let us consider X denotes the number of bulbs that fuse after 150 days.

Now, a binomial distribution X follows with n = 5.

Here p = 0.05 and q = 0.95.

Or p = 1/20 and q = 19/20

P(X = r) = $^{5}{}{C}_r \left( \frac{1}{20} \right)^r \left( \frac{19}{20} \right)^{5 - r}$

So, the probability (not more than 1 will fuse after 150 days of use) is

P(X < 1) = P(X = 0) + P(X = 1)

= $\left( \frac{19}{20} \right)^5 + 5 C_1 \left( \frac{1}{20} \right)^1 \left( \frac{19}{20} \right)^{5 - 1}$

= $\left( \frac{19}{20} \right)^4 \left\{ \frac{19}{20} + \frac{5}{20} \right\}$

= (6/5) (19/20)⁴

(iii) 找出 5 个此类灯泡中超过一个在使用 150 天后熔断的概率

解决方案：

Let us consider X denotes the number of bulbs that fuse after 150 days.

Now, a binomial distribution X follows with n = 5.

Here p = 0.05 and q = 0.95

Or p = 1/20 and q = 19/20

P(X = r) = $^{5}{}{C}_r \left( \frac{1}{20} \right)^r \left( \frac{19}{20} \right)^{5 - r}$

So, the probability(more than one will fuse after 150 days of use) is

P(X > 1) = 1 – P(X < 1)

= 1 – (6/5) (19/20)⁴

(iv) 求 5 个这样的灯泡中至少有一个在使用 150 天后熔断的概率。

解决方案：

Let us consider X denotes the number of bulbs that fuse after 150 days.

Now, a binomial distribution X follows with n = 5.

Here p = 0.05 and q = 0.95

Or p = 1/20 and q = 19/20

P(X = r) = $^{5}{}{C}_r \left( \frac{1}{20} \right)^r \left( \frac{19}{20} \right)^{5 - r}$

So, the probability(at least one will fuse after 150 days of use) is

P(X > 1) = 1 – P(X = 0)

= 1 – (19/20)⁵

问题 18。假设 90% 的人是右撇子。 10 个随机样本中最多有 6 个是右撇子的概率是多少？

解决方案：

Let us consider X denotes the number of people that are right-handed in the sample of 10 people.

Now, a binomial distribution X follows with n = 10.

Here p = 90 % = 90/100 = 0.9

And q = 1 – p = 0.1

P(X = r) = ¹⁰C_r (0.9)^r (0.1)^10-r

So, the probability that at most 6 are right – handed is

P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)

= 1 – {P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)}

= $1 - \sum^{10}_{r = 7}{^{10}{}{C}_r} (0 . 9 )^r (0 . 1 )^{10 - r}$