Question

Рабочий обслуживает 1000 станков. Вероятность сбоя на одном 0,004. Найти вероятность того что сбой произойдет на пяти станках.

Answer 1

1)5*0,004=0,02- вероятнось сбоя на 5 станках 2)1000*0,004=4 вероятность сбоя на 1000 3)0,02/4=0,005 -вероятнось того, что сбой произойдет на 5 станках

Answer 2

Calculation of the Probability of Failure on Five Machines

To find the probability that a failure will occur on five machines, we can use the binomial distribution formula. The binomial distribution is used to calculate the probability of a certain number of successes in a fixed number of independent Bernoulli trials, where each trial has the same probability of success.

In this case, the probability of a failure on one machine is given as 0.004, and the worker is servicing 1000 machines. We want to find the probability of a failure occurring on five machines.

The binomial distribution formula is as follows:

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 on a single trial - n is the number of trials

In this case, we want to find the probability of getting exactly 5 failures, so k=5.

Let's calculate the probability using the given values:

n = 1000 (number of machines) p = 0.004 (probability of failure on one machine) k = 5 (number of failures)

Using the binomial distribution formula, we can calculate the probability as follows:

P(X=5) = C(1000, 5) * (0.004)^5 * (1-0.004)^(1000-5)

Now, let's calculate the probability using the formula:

P(X=5) = (1000! / (5! * (1000-5)!)) * (0.004)^5 * (1-0.004)^(1000-5)

P(X=5) = (1000! / (5! * 995!)) * (0.004)^5 * (0.996)^995

P(X=5) = (1000 * 999 * 998 * 997 * 996 / (5 * 4 * 3 * 2 * 1)) * (0.004)^5 * (0.996)^995

P(X=5) = 0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000#### Calculation of the Probability of Failure on Five Machines

To calculate the probability of a failure occurring on five machines, we can use the binomial distribution formula. The binomial distribution is used to calculate the probability of a specific number of successes (in this case, failures) in a fixed number of independent trials.

Given that the probability of a failure on one machine is 0.004, we can use the binomial distribution formula to calculate the probability of exactly five failures in 1000 trials (machines).

The formula for the binomial distribution is:

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

Where: - P(X=k) is the probability of getting exactly k failures - C(n, k) is the number of combinations of n items taken k at a time - p is the probability of a single failure - n is the number of trials (machines)

In this case, we want to calculate P(X=5), where p = 0.004 and n = 1000.

Plugging in the values into the formula, we get:

P(X=5) = C(1000, 5) * (0.004)^5 * (1-0.004)^(1000-5)

Now, let's calculate the probability using the formula.

Calculation:

Using a calculator or a statistical software, we can calculate the probability of exactly five failures in 1000 trials (machines) as follows:

P(X=5) ≈ 0.135

Therefore, the probability of a failure occurring on five machines is approximately 0.135, or 13.5%.

Please note that this calculation assumes that the failures on each machine are independent events and that the probability of failure remains constant for each machine.