Question

Опытом установлено, что в среднем 70% массовой продукции, выпускаемой некоторой мастерской, принадлежит первому сорту. Какова вероятность того, что из 6 взятых наудачу изделий этой мастерской окажется не менее пяти изделий первого сорта.

Answer 1

Вероятность первого сорта  - р = 0,7 - дано
Вероятность НЕ первого сорта  q = 1 - 0.7 = 0.3 
Вероятность шести событий из 5 ДА и одного НЕТ вычисляется по формуле
Р(A) = p⁵q = 0.7*0.7*0.7*0.7*0.7*0.3 = 0.7⁵*0.3 = 0.16807 * 0.3 ~ 0.050=5% - ОТВЕТ
Так мало, потому, что все пять из шести будут первого сорта.

Answer 2

Problem Analysis

We are given that, on average, 70% of the mass-produced items from a certain workshop belong to the first grade. We need to find the probability that out of 6 randomly selected items from this workshop, at least 5 will be of the first grade.

Solution

To solve this problem, we can use the binomial probability formula. The formula for the probability of getting exactly k successes in n trials, where the probability of success in a single trial is p, is given by:

P(X = k) = C(n, k) * p^k * (1-p)^(n-k)

Where: - P(X = k) is the probability of getting exactly k successes - C(n, k) is the number of combinations of n items taken k at a time - p is the probability of success in a single trial - (1-p) is the probability of failure in a single trial - n is the total number of trials

In this case, the probability of success (getting an item of the first grade) in a single trial is 0.7, and the total number of trials is 6.

To find the probability of getting at least 5 items of the first grade, we need to calculate the probabilities of getting exactly 5, 6 items of the first grade and sum them up.

Calculation

Let's calculate the probability using the formula mentioned above:

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

P(X = 5) = C(6, 5) * 0.7^5 * (1-0.7)^(6-5)

P(X = 6) = C(6, 6) * 0.7^6 * (1-0.7)^(6-6)

Answer

Calculating the probabilities:

P(X = 5) = 6 * 0.7^5 * (1-0.7)^(6-5) = 0.302526

P(X = 6) = 1 * 0.7^6 * (1-0.7)^(6-6) = 0.117649

Summing up the probabilities:

P(X >= 5) = P(X = 5) + P(X = 6) = 0.302526 + 0.117649 = 0.420175

Therefore, the probability that out of 6 randomly selected items from this workshop, at least 5 will be of the first grade is approximately 0.420175 or 42.02%.

Please note that the above calculations are based on the assumption that the probability of getting an item of the first grade is constant for each trial and that the trials are independent of each other.