在某所学生人数众多的学校中,有 p%p\%(其中 p>40p > 40)的学生学习 H2 生物。

  1. 在 40 名学生的随机样本中,学习 H2 生物的人数不少于 9 且不多于 20 的概率为 0.25。求 pp 的值。

  2. 已知 p=60p = 60。一组随机抽取的 12 名学生将在会议室参加面试,每次只面试一人。求在面试的第 3 位和第 10 位 student 分别是他们之中第 1 个和第 4 个学习 H2 生物的学生的概率。

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完整解法

7[0]

1. Define the binomial random variable for the number of students studying H2 Biology.

Let XX be the number of students studying H2 Biology in the sample of 40.
X∼B(40,p100)X \sim \text{B}\left(40, \frac{p}{100}\right)

2. Formulate the equation based on the given probability.

We are given that the probability of no less than 9 but no more than 20 students studying H2 Biology is 0.250.25.
P(9≤X≤20)=0.25P(9 \le X \le 20) = 0.25
P(X≤20)−P(X≤8)=0.25P(X \le 20) - P(X \le 8) = 0.25

3. Solve the equation for $p$ using a graphing calculator.

Using a graphing calculator, we can graph Y1=P(X≤20)−P(X≤8)−0.25Y_1 = P(X \le 20) - P(X \le 8) - 0.25 where the probability of success is x=p100x = \frac{p}{100}.
We find the root for x>0.4x > 0.4 (since it is given that p>40p > 40):
x≈0.5653x \approx 0.5653
p100=0.5653\frac{p}{100} = 0.5653
p=56.5 (to 3 s.f.)p = 56.5 \text{ (to 3 s.f.)}

答案

56.556.5

7[1]

1. Find the probability for the specific sequence of the first three interviewees.

Let SS denote a student who studies H2 Biology, and S′S' denote a student who does not.
Given p=60p = 60, the probability of a randomly chosen student studying H2 Biology is P(S)=0.6P(S) = 0.6, and P(S′)=0.4P(S') = 0.4.
For the 3rd interviewee to be the 1st student studying H2 Biology, the 1st and 2nd interviewees must be S′S', and the 3rd must be SS.
Probability=0.4×0.4×0.6=0.096\text{Probability} = 0.4 \times 0.4 \times 0.6 = 0.096

2. Find the probability for the 4th to 10th interviewees satisfying the conditions.

For the 10th interviewee to be the 4th student studying H2 Biology, there must be exactly 2 students studying H2 Biology among the 6 interviewees from the 4th to 9th positions, and the 10th interviewee must be SS.
Probability=(62)(0.6)2(0.4)4×0.6\text{Probability} = \binom{6}{2} (0.6)^2 (0.4)^4 \times 0.6
=15×0.36×0.0256×0.6=0.082944= 15 \times 0.36 \times 0.0256 \times 0.6 = 0.082944

3. Calculate the total probability of the entire sequence.

Since the selections of interviewees are independent, we multiply the probabilities of the required sequences for the first 10 interviewees. The outcomes of the 11th and 12th interviewees do not affect this condition.
Required probability=0.096×0.082944\text{Required probability} = 0.096 \times 0.082944
=0.007962624= 0.007962624
≈0.00796 (to 3 s.f.)\approx 0.00796 \text{ (to 3 s.f.)}

答案

0.007960.00796

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