Question

Производители деталей выявили, что на одной из сборочных линий 35% производимых деталей имеют дефект. Во время очередной проверки инспектор выбирает 7 деталей из этой сборочной линии. Найдите вероятность того, что инспектору попадутся две детали с дефектом. Ответ округлите до тысячных.

Answer 1

Р(А) = n/m

n=2

m=7

P(A)=2/7

Answer 2

Calculation of Probability

To find the probability that the inspector will select two defective parts out of the seven parts chosen, we need to use the concept of combinations.

The probability of selecting two defective parts can be calculated using the formula:

P = (C(n, k) * C(N-n, K-k)) / C(N, K)

Where: - P is the probability of selecting two defective parts - C(n, k) is the number of ways to choose k items from a set of n items - N is the total number of parts on the assembly line - n is the number of defective parts on the assembly line - K is the number of parts chosen by the inspector - k is the number of defective parts chosen by the inspector

In this case: - N is the total number of parts on the assembly line, which is not given in the question. - n is the number of defective parts on the assembly line, which is 35% of the total number of parts on the assembly line. - K is the number of parts chosen by the inspector, which is 7. - k is the number of defective parts chosen by the inspector, which is 2.

Since the total number of parts on the assembly line is not given, we cannot calculate the exact probability. We need the total number of parts on the assembly line to proceed with the calculation.

Please provide the total number of parts on the assembly line so that we can calculate the probability.