Question

(Алгебра событий. Основные теоремы теории вероятностей) В бригаде 3 трактора которые исправны с вероятностями 0,7; 0,8; 0,9соответственно. Какова вреоятность того, что на день проверки только два трактора окажутся исправными?

Answer 1

Вероятность события( исправны только два трактора)=== (исправны 1 и 2, неисправен 3) +(исправны 1 и 3, неисправен 2)+(исправны 2 и 3, неисправен 1) = 0,7*0,8*0,1 + 0,7*0,9*0,2 + 0,8*0,9*0,3 = 0,056 + 0,126+0,216 =0,398.

Answer 2

Algebra of Events and Basic Theorems of Probability

In a brigade, there are 3 tractors, each with respective probabilities of being operational: 0.7, 0.8, and 0.9. We want to find the probability that only two tractors will be operational on the day of inspection.

To solve this problem, we can use the concept of the complement of an event and the addition rule of probability.

The probability of exactly two tractors being operational can be calculated using the binomial probability formula:

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

Where: - n = total number of tractors - k = number of tractors that are operational - p = probability of a tractor being operational - (n choose k) = n! / (k! * (n-k)!)

Let's calculate the probability step by step.

First, we need to find the probability of exactly two tractors being operational.

The probability of the first two tractors being operational and the third one failing is: P(0.7 * 0.8 * (1-0.9))

The probability of the first and third tractors being operational and the second one failing is: P(0.7 * (1-0.8) * 0.9)

The probability of the second and third tractors being operational and the first one failing is: P((1-0.7) * 0.8 * 0.9)

Now, we can use the addition rule of probability to find the total probability of only two tractors being operational: P(2 operational) = P(0.7 * 0.8 * (1-0.9)) + P(0.7 * (1-0.8) * 0.9) + P((1-0.7) * 0.8 * 0.9)

Let's calculate the final probability.