Question

на заводе 30% произведённых подшипников имеют дефект при контроле качества продукции выявляется 65 дефектных подшипников остальные подшипники поступают в продажу Найдите вероятность того что случайно выбранная при покупке подшипник не имеет дефектов Ответ округлите до сотых​

Answer 1

Ответ:

Объяснение:

Пусть X - количество произведенных подшипников.

Тогда 0,7X - количество подшипников, не имеющих дефект.

0,3X - количество подшипников с дефектом до контроля качества.

0,3*(1-0,65)X = 0,105X - количество дефектных подшипников, поступивших в продажу после контроля качества.

Тогда вероятность того, что случайно выбранный при покупке подшипник не имеет дефектов равна:

P = 0,7X/(0,105X+0,7X) = 0,7/0,805 = 0,8695...

Округляя до сотых, получаем 0,87.

Answer 2

Calculation of Probability

To find the probability that a randomly selected bearing does not have defects, we need to consider the information provided.

According to the given information: - 30% of the bearings produced have defects. - During quality control, 65 defective bearings are identified. - The remaining bearings are sold.

To calculate the probability, we can use the following formula:

Probability of a non-defective bearing = (Number of non-defective bearings) / (Total number of bearings)

To find the number of non-defective bearings, we need to subtract the number of defective bearings from the total number of bearings produced.

Let's assume that the total number of bearings produced is represented by 'x'.

The number of defective bearings is 30% of 'x', which is 0.3x.

Therefore, the number of non-defective bearings is 'x' minus 0.3x, which is 0.7x.

The total number of bearings is the sum of the defective and non-defective bearings, which is 0.3x + 0.7x = x.

Now, we can calculate the probability:

Probability of a non-defective bearing = (0.7x) / x = 0.7

Therefore, the probability that a randomly selected bearing does not have defects is 0.7 or 70%.

Please note that the answer has been rounded to the nearest hundredth as requested.