第 12 类 RD Sharma 解决方案 - 第 33 章二项分布 - 练习 33.2 |设置 2

问题 15. 掷三次骰子。一次投掷 1 或 6 就是成功。求成功次数的均值和方差。

解决方案：

Let p denote the success and q denote failure of an event.

Now, the sample space when a dice is thrown is given by S = {1, 2, 3, 4, 5, 6}

Hence, p = 2/6 = 1/3 and q = 1 – 1/3 = 2/3

Therefore, Mean = np = 3 × 1/3 = 1 and Variance = npq = 1 × 2/3 = 2/3

问题 16. 如果随机变量 X 服从均值为 3，方差为 3/2 的二项分布，求 P (X ≤ 5)。

解决方案：

We are given mean (np) = 3 and variance (npq) = 3/2.

Solving for the value of q, $\frac{npq}{np} = \frac{\frac{3}{2}}{3}$

q = 1/2, hence we can conclude p = 1 – 1/2 = 1/2

Now putting the value of p in relation, np = 3, we get n = 6

We know that a binomial distribution follows the relation:

P(X = r) = ⁿC_r p^r(q)^n-r

Therefore, in this case P(X = r) = ⁶C_r (1/2)^r(1/2)^6-r 

P(X = r) = ⁶C_r (1/2)⁶  

We are required to calculate the value for P(X ≤ 5) = 1 – P(X = 6)

P(X ≤ 5) = 1 – ⁶C_r (1/2)⁶ 

P(X ≤ 5) = 1 – (1/64) 

P(X ≤ 5) = 63/64 

问题 17.如果 X 服从均值为 4，方差为 2 的二项分布，求 P(X ≥ 5)。

解决方案：

We are given mean (np) = 4 and variance (npq) = 2.

Solving for the value of q, npq/np = 2/4 

q = 1/2, hence we can conclude p = 1 – 1/2 = 1/2

Now putting the value of p in relation, np = 4, we get n = 8

We know that a binomial distribution follows the relation: P(X = r) = ⁿC_r p^r(q)^n-r

Therefore, in this case P(X = r) = ⁸C_r (1/2)^r(1/2)^8-r  

P(X = r) = ⁸C_r (1/2)⁸  

We are required to calculate the value 

P(X ≥ 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)

P(X ≥ 5) = ⁸C₅ (1/2)⁸ + ⁸C₆ (1/2)⁸ + ⁸C₇ (1/2)⁸ + ⁸C₈ (1/2)⁸  

P(X ≥ 5) = (1/2)⁸[⁸C₅ + ⁸C₆ + ⁸C₇ + ⁸C₈]

P(X ≥ 5) = (56 + 28 + 8 + 1)/256

P(X ≥ 5) = 93/256 

问题 18.二项分布的均值和方差分别为 4/3 和 8/9。求 P(X ≥ 1)。

解决方案：

We are given mean (np) = 4 and variance (npq) = 2

Solving for the value of q, npq/np = $\frac{\frac{8}{9}}{\frac{4}{3}}$

q = 2/3, hence we can conclude p = 1 – 2/3 = 1/3

Now putting the value of p in relation, np = 4/3, we get n = 4

We know that a binomial distribution follows the relation: P(X = r) = ⁿC_r p^r(q)^n-r

Therefore, in this case P(X = r) = ⁴C_r (1/3)^r(2/3)^4-r  

We are required to calculate the value for P(X ≥ 1) = 1 – P(X = 0)

P(X ≥ 1) = 1 – ⁴C₀ (1/3)⁰(2/3)⁴  

P(X ≥ 1) = 1 – 16/81

P(X ≥ 1) = 65/81

问题 19.如果 6 次试验的二项式分布的均值和方差之和为 10/3，求分布。

解决方案：

Given n = 6 and np + npq = 10/3

np (1 + q) = 10/3

6p (1 + 1 – p) = 10/3

12p – 6p² = 10/3

18p² – 36p + 10 = 0

Solving for the value of p we will get p = 1/3 or p = 5/3. 

Since, the value of p cannot exceed 1, we will consider p = 1/3.

Therefore, q = 1 – 1/3 = 2/3

Now, a binomial distribution is given by the relation: ⁿC_r p^r(q)^n-r

P(x = r) = ⁶C_r (1/3)^r(2/3)^6-r for r = 0,1,2,….,6

问题 20.一对骰子被掷了 4 次。如果获得双峰被认为是成功的，请找到成功次数的概率分布，从而找到它的均值。

解决方案：

We are given n = 4 and 

a doublet in the throw of a dice occurs when we get (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)

Therefore, the probability of success, p = 6/36 = 1/6, so q = 1 – 1/6 = 5/6

Now, a binomial distribution is given by the relation: ⁿC_r p^r(q)^n-r

P(x = r) = ⁴C_r (1/6)^r(5/6)^4-r for r = 0, 1, 2, 3, 4

Hence, the probability distribution is given as:

X	0	1	2	3	4
P(X)	625/1296	500/1296	150/1296	20/1296	1/1296

Mean = 0 × (625/1296) + 1 × (500/1296) + 2 × (150/1296) + 3 × (20/1296) + 0 × (1/1296)

= 864/ 1296

= 2/3

问题 21.求一对骰子投掷 3 次中双骰数的概率分布，并求其均值。

解决方案：

We are given n = 3 and 

a doublet in the throw of a dice occurs when we get (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)

Therefore, the probability of success, p = 6/36 = 1/6, so q = 1 – 1/6 = 5/6

Now, a binomial distribution is given by the relation: ⁿC_r p^r(q)^n-r

P(x = r) = ³C_r (1/6)^r(5/6)^3-r for r = 0, 1, 2, 3

Hence, the probability distribution is given as:

X	0	1	2	3
P(X)	125/216	75/216	15/216	1/216

Mean = 0 × (125/216) + 1 × (75/216) + 2 × (15/216) + 3 × (1/216) 

= 108/216

= 1/2

问题 22.从包含 5 个缺陷的 15 个灯泡中，一个一个地抽取 4 个灯泡的样品并更换。求缺陷灯泡数量的概率分布。因此，求分布的均值。

解决方案：

Total number of bulbs = 15 and total defective bulbs = 5 

Thus, the probability of getting one defective bulb with replacement, p = 5/15 = 1/3

 Hence, q = 1 – 1/3 = 2/3.

Now, a binomial distribution is given by the relation: ⁿC_r p^r(q)^n-r

P(x = r) = ⁴C_r (1/3)^r(2/3)^4-r for r = 0, 1, 2, 3, 4

Hence, the probability distribution is given as:

X	0	1	2	3	4
P(X)	16/81	32/81	24/81	8/81	1/81

Mean = 0 × (16/81) + 1 × (32/81) + 2 × (24/81) + 3 × (8/81) + 4 × (1/81) 

= 108/81

= 4/3

问题 23. 掷骰子 3 次。令 X 为“看到的二的数量”。求 X 的期望。

解决方案：

We are given the number of throws, n = 3

Let p denote the probability of getting a 2 in the throw of a dice, then p = 1/6

Therefore, we can conclude 1 = 1 – 1/6 = 5/6

Now, the expectation of X denotes mean therefore, E(X) = np = 3 × 1/6 = 1/2

问题 24.掷骰子两次。 “成功”是在掷骰中得到偶数。求成功次数的方差。

解决方案：

We are given the number of times the coin is tossed, n = 2

Let p denote the probability of getting even number on dice upon throwing which is a success.

Thus, p = 3/6 = 1/2, therefore we can conclude q = 1 – p = 1 – 1/2 = 1/2

Now, the variance is given by npq.

Variance = 2 × 1/2 × 1/2 = 1/2

问题 25. 从一副洗好的 52 张牌中依次抽出三张牌并替换。求黑桃数量的概率分布。因此，求分布的均值。

解决方案：

Number of cards drawn with replacement, n = 3

p = Probability of getting a spade card upon withdrawal = 13/52 = 1/4

Thus, we can conclude, q = 1 – 1/4 = 3/4

Now, a binomial distribution is given by the relation: ⁿC_r p^r(q)^n-r

P(x = r) = ³C_r (1/4)^r(3/4)^3-r for r = 0, 1, 2, 3 

Hence, the probability distribution is given as:

X	0	1	2	3
P(X)	27/64	27/64	9/64	1/64

Mean = 0 × (27/64) + 1 × (27/64) + 2 × (9/64) + 3 × (1/64) 

= (27 + 18 + 3)/64

= 48/64 

= 3/4

问题 26.一个瓮中有 3 个白球和 6 个红球。四个球一一抽出，并从瓮中更换。求出红球数的概率分布。另外，求分布的均值和方差。

解决方案：

Let p denote the probability of drawing a red ball which is considered a success, p = 6/9 = 2/3

And the probability of drawing a white ball which is considered a failure, q = 3/9 = 1/3

We have to draw four balls, so n = 4.

Hence, the mean of the probability distribution = np = 4 × 2/3 = 8/3

And variance = npq = 8/3 × 1/3 = 8/9

Now, a binomial distribution is given by the relation: ⁿC_r p^r(q)^n-r

P(x = r) = ⁴C_r (2/3)^r(1/3)^4-r for r = 0, 1, 2, 3, 4

Hence, the probability distribution is given as:

X	0	1	2	3	4
P(X)	1/81	8/81	24/81	32/81	16/81

问题 27. 5 个坏橙子不小心与 20 个好橙子混在一起。如果四个橙子一个接一个地被放回，则求出坏橙子数量的概率分布。因此，找到分布的均值和方差。

解决方案：

Let p denote the probability of drawing a bad orange which is considered a success, p = 5/25 = 1/5

And the probability of drawing a good orange which is considered a failure, q = 20/25 = 4/5

We have to draw four oranges, so n = 4

Hence, the mean of the probability distribution = np = 4 × 1/5 = 4/5

And the variance = npq = 4/5 × 4/5 = 16/25

Now, a binomial distribution is given by the relation: ⁿC_r p^r(q)^n-r

P(x = r) = ⁴C_r (1/5)^r(4/5)^4-r for r = 0, 1, 2, 3, 4

Hence, the probability distribution is given as:

X	0	1	2	3	4
P(X)	256/625	256/625	96/625	16/625	1/625