Question

В магазине выставлены для продажи n изделий, среди которых k изделий некачественные. Какова вероятность того, что взятые случайным образом m изделий будут некачественными n=18, k=6, m=3

Answer 1

Вероятность того, что m = 2 изделия, которые вытащили из партии n = 10 изделий, где k = 4 изделия некачественные (качественных 10 - 4 = 6 изделий), окажутся некачественными можно рассчитать по формуле:

Р = С06 * С24 / С210.

Сочетания равны:

С06 = 6! / 0! * (6 - 0)! = 6!/ (0! * 6!) = 1.

С24 = 4! / 2! * (4 - 2)! = 4!/ (2! * 2!) = 6.

С210 = 10! / 2! * (10 - 2)! = 10!/ (2! * 8!) = 45.

Рассчитаем вероятность:

Р = С06 * С24 / С210 = 1 * 6 / 45 = 0,133 = 13,3%.

Ответ: Вероятность равна 0,133 или 13,3%.

Answer 2

Calculation of Probability

To calculate the probability of randomly selecting m defective items out of n items, we can use the concept of combinations. The probability can be calculated using the formula:

P(m defective items) = (C(k, m) * C(n-k, m)) / C(n, m)

Where: - n is the total number of items available for sale - k is the number of defective items among the n items - m is the number of items randomly selected

In this case, n = 18, k = 6, and m = 3. Let's calculate the probability using the formula.

Calculation

Using the formula mentioned above, the probability of randomly selecting 3 defective items out of 18 items, with 6 defective items among them, can be calculated as follows:

P(3 defective items) = (C(6, 3) * C(18-6, 3)) / C(18, 3)

P(3 defective items) = (C(6, 3) * C(12, 3)) / C(18, 3)

Now, let's calculate the combinations:

C(6, 3) = 6! / (3! * (6-3)!) = 6! / (3! * 3!) = (6 * 5 * 4) / (3 * 2 * 1) = 20

C(12, 3) = 12! / (3! * (12-3)!) = 12! / (3! * 9!) = (12 * 11 * 10) / (3 * 2 * 1) = 220

C(18, 3) = 18! / (3! * (18-3)!) = 18! / (3! * 15!) = (18 * 17 * 16) / (3 * 2 * 1) = 816

Now, let's substitute the values into the probability formula:

P(3 defective items) = (20 * 220) / 816 ≈ 0.537

Therefore, the probability of randomly selecting 3 defective items out of 18 items, with 6 defective items among them, is approximately 0.537.

Please note that the calculation assumes that the selection of items is truly random and that the defective items are evenly distributed among the available items.