伯努利试验和二项式分布–概率

在本文中，我们将与相关定理一起详细讨论伯努利试验和二项分布。伯努利审判也被称为二项式审判。在伯努利试验的情况下，只有两个可能的结果，但是在二项式分布的情况下，我们通过一系列独立实验获得成功的次数。

伯努利的审判

让我们考虑随机实验E的n个独立重复(试验)。如果A是与E相关的事件，使得重复的P(A)保持不变，则这些试验称为伯努利试验。

定理：如果在伯努利实验的单次试验中某事件发生的概率(成功概率)为p，则在n次独立试验中该事件恰好发生r次的概率等于ⁿ C _r q ^{n – r} p ^r ，其中q = 1 – p ，该事件失败的概率。

In Short: 

Required Probability = ⁿC_r q^{n – r} p^r

where, 

p = Probability of Success

q = 1 – p = Probability of Failure

n = Number of Independent trials

r = The number of times an event occurred 

证明：

Getting exactly r successes means getting r successes and (n – r) failures simultaneously.

∴ P(getting r successes and  n – r failures) = q^{n – r} p^r (since the n trials are independent) [By Product Theorem]

The trials, from which the successes are obtained, are not specified. There are  ⁿC_r ways of  choosing r trials for successes. Once the r trials are chosen for successes, the remaining  (n – r) trials should result in failures.These  ⁿC_r  ways are mutually exclusive. In each of these ⁿC_r  ways, P(getting exactly r successes) = q^{n – r} p^r

Therefore, by the addition theorem, the required probability = ⁿC_r q^{n – r} p^r

伯努利定理的推广

多项式分布：

If  A₁, A₂, . . . , A_k  are exhaustive and mutually exclusive events associated with a random experiment such that, P(A_i occurs ) = p_i where, 

p₁ + p₂ +. . . + p_k = 1, and if the experiment is repeated n times, then the probability A₁ occurs r₁ times, A₂ occurs r₂ times,. . . . ,A_k occurs r_k times is given by:

P_n(r₁, r_{2 , . . . ,} r_k) = $\frac{n!}{r_{1}! r_{2}! . . . r_{k}!} \ p_{1}^{r_{1}}\times p_{2}^{r_{2}}\times. . .\times p_{k}^{r_{k}}$

where, 

r₁​ + r₂ ​+ …+ r_k ​= n    

证明：

The r₁ trials in which the event A₁ occurs can be chosen from the n trials ⁿC_r ways. The remaining (n – r₁) trials are left over for the other events. 

The r₂ trials in which the event A₂  occurs can be chosen from the (n – r₁) trials in ^{(n – r₁)}C_r2  ways.

The r₃ trials in which the event A₃ occurs can be chosen from the (n – r₁ – r₂) trials in ^{(n – r₁ – r₂)}C_r3 ways, and so on.

Therefore the number of ways in which the events A₁​, A₂​, …, A_k can happen:

ⁿ​C_r1​ ​× ^{(n − r₁​)}C_r2 ​​× ^{(n −r₁ ​− r₂​)}C_r3​​ × ^{(n−r₁​ − r₁​ – …− r_k ​− 1)}C_rk​​ =  n!/(r₁!r₂! . . . r₃!)

Consider any one of the above ways in which the events A₁, A₂, . . ., A_k occur.

Since the n trials are independent, r₁ + r₂ + . . . +r_k trials are also independent.

∴ P(A₁ occurs r₁ times, A₂ occurs r₂ times, . . . , A _k  occurs r _k times) = p₁ ^r₁ × p₂^r₂  ×  . . . × p_k ^r_k 

Since the ways in which the events happen are mutually exclusive, the required probability is given by

P_n (r₁ , r₂ , . . . , r _k ) =$\frac{n!}{r_{1}! r_{2}! . . . r_{k}!}$[Tex]\times \ p_{1} ^{r_{1}}\times p_{2}^{r_{2}}\times… \times p_{k}^{r_{k}}[/Tex]

例子

例1：一枚硬币被扔了无数次。如果头的在一个单一的抛的概率为p，表明第k个头在第n个辗转反侧获得，但概率不早为^{(n -1)C} _{K -1} ^pk信息将q^{个n – k，}其中q = 1 – p。

解决方案：

K heads should be obtained at the nth tossing, but not earlier.

Therefore, (k – 1) heads must be obtained in the first (n – 1) tosses and 1 head must be obtained at the nth toss.

Required probability = P[ k – 1 heads in (n – 1) tosses]  ×  P(1 head in 1 toss)]  = ⁽ⁿ⁻¹⁾C_k−1​p^k-1q^n−k x p 

示例2：如果在一个有2个孩子的家庭中至少有1个孩子是男孩，那么两个孩子都是男孩的概率是多少?

解决方案：

p = Probability that a child is a boy = 1/2.

∴ q = 1/2 and n = 2

P(at least one boy) = p (exactly 1 boy) + p (exactly 2 boys) = $2C_{1}\frac{1}{2}^2 + 2C_{2}\frac{1}{2}^2$= $\frac{3}{4}$

∴ Required probability  = P(both are boys) / P( at least 1 boy) = $\frac{1}{4}/\frac{3}{4} = \frac{1}{3}$ 

二项分布

定理：令A为与随机实验E相关的事件，使得P(A)= p且P(A’)= q = 1 – p。假设p对于所有重复都保持相同，如果我们考虑E的n个独立重复(或试验)，并且随机变量(RV)X表示事件A发生的次数，则X称为二项式随机变量，其中参数n和p，或者我们可以说X遵循参数n和p的二项式分布，或者符号B(n，p)。显然，X可以取的可能值为0、1、2，…，n。根据伯努利试验的定理，二项式RV的概率质量函数由下式给出：

P_{(x = r)} = ⁿC_r q^{n – r} p^r, r = 0, 1, 2, …, n

where,

p + q = 1

笔记：

1. Binomial distribution is a legitimate probability distribution since

$\sum_{r=0}^n P(X=r)=\sum_{r=0}^n nCrq^{n-r}p^r=(q+p)^n=1$

2. The  Mean of the Binomial Distribution is given by: $E(x)=\sum_{r} x_{r}p_{r}=np$; also $E(x^2)=\sum_{r}x_{r}^2p_{r}$

3.The Variance of the Binomial Distribution is given by: $Var(x)=E(x^2)-{{E(x)}}^2=npq$

例子

例1：在每个有4个孩子的800个家庭中，预计会有多少个家庭

(i)2个男孩和2个女孩，

(ii)至少1个男孩，

(iii)最多2个女孩，

(iv)两性的孩子。

假设男孩和女孩的概率相等。

解决方案：

Considering each child as a trial, n = 4.Assuming that birth of a boy is a success, p = 1/2, and q = 1/2. Let X denote the number of successes(boys).

(i) P(2 boys and 2 girls) = P(X = 2)

                                    = $4C_{2} \times (\frac{1}{2})^2 \times (\frac{1}{2})^{4-2}=6 \times(\frac{1}{2})^4=\frac{3}{8}$           

$\therefore$ No. of  families having 2 boys and 2 girls

                                    = N.P(X = 2) [N is the total no. of families considered]

                                    = $800 \times \frac{3}{8}=300$

(ii) P(at least 1 boy) = P(X ≥ 1)

                               = P(X = 1) + P( X = 2) + P(X = 3) + P(X = 4)

                               = 1 – P(X = 0)

                               = $1-4C_{0} \times(\frac{1}{2})^0\times(\frac{1}{2})^4=1-\frac{1}{16}=\frac{15}{16}$ 

$\therefore$ No. of families having at least 1 boy

                               = $800 \times \frac{15}{16}=750$ 

(iii) P(at most 2 girls) = P(exactly 0 girl, 1 girl or 2 girls)

                                 = P(X = 4, X = 3 or X = 2)

                                 = $1-(4C_{0} \times (\frac{1}{2})^4+4C_{1} \times(\frac{1}{2})^4)=\frac{11}{16}$

$\therefore$ No of families having at most 2 girls

                                 = $800 \times \frac{11}{16}=550$

(iv) P(children of both sexes)

                                 = 1 – P( children of same sex )

                                 = 1 – {P(all are boys) + P(all are girls)}

                                 = 1 – {P(X = 4) + P(X = 0)}

                                 = $1-(4C_{4} \times(\frac{1}{2})^4 + 4C_{0}\times(\frac{1}{2})^4)=1-\frac{1}{8}=\frac{7}{8}$

$\therefore$ No of families having children of both sexes

                                 = $800\times\frac{7}{8} = 700$

示例2：同时投掷十枚硬币。找出获得至少七个脑袋的可能性吗?

解决方案：

p = Probability of getting a head = 1/2

q = Probability of not getting a head = 1 – p = 1/2

The probability of getting x heads in a random throw of 10 coins is: 

p(x) = $10C_{x}\times\frac{1}{2}^x\times\frac{1}{2}^{10-x}=10C_{x}\times\frac{1}{2}^{10}$; x = 0, 1, 2, . . . , 10

$\therefore$ Probability of getting at least seven heads is given by :

P(X ≥ 7) = p(7) + p(8) + p(9) + p(10)

              = $\frac{1}{2}^{10}( 10C_{7} + 10C_{8} + 10C_{9} + 10C_{10}) =\frac{176}{1024}$