Question

Оптовая база обслуживает десять предприятий. От каждого из них может поступить заявка на товары на текущий день с вероятностью 0,3, независимо от заявок других предприятий. Найти наивероятнейшее число заявок в день и вероятность получения этого числа заявок.

Answer 1

N=12 p=0.3 q= 1-p=0.7 Наивероятнейшее число k: np-q <= k < np+p 3,6-0,7 <= k < 3,6+0,3 2,9 <= k < 3,9 Откуда k=3 Ну а теперь находи по формуле Бернулли Р (k=3) при n=12, p=0.3, q=1-p=0.7, k=3 Р (12,3)=С (12,3)* p^3* q^9 Подсчитай и проверь С - сочетания

Answer 2

Naive Probability Calculation

To find the naively most probable number of requests per day and the probability of receiving that number of requests, we can use the information provided.

Given that each of the ten enterprises can submit a request with a probability of 0.3 independently of the others, we can use the binomial distribution to calculate the most probable number of requests and its probability.

Calculation

The most probable number of requests can be calculated using the formula:

np = n * p

Where: - n = number of enterprises (10) - p = probability of a request from a single enterprise (0.3)

np = 10 * 0.3 = 3

So, the most probable number of requests per day is 3.

The probability of receiving this number of requests can be calculated using the binomial probability formula:

P(x) = (nCx) * (p^x) * ((1-p)^(n-x))

Where: - n = number of trials (10 enterprises) - x = number of successful trials (3 requests) - p = probability of success on an individual trial (0.3)

P(3) = (10C3) * (0.3^3) * ((1-0.3)^(10-3))

Using the binomial coefficient (nCk) formula, we can calculate (10C3) = 120.

P(3) = 120 * (0.3^3) * (0.7^7) ≈ 0.2668

So, the probability of receiving 3 requests per day is approximately 0.2668.

Therefore, the naively most probable number of requests per day is 3, with a probability of approximately 0.2668.