Question

В ящике 15 деталей, из которых 9 окрашенных. Наудачу извлечены 4 детали. Найти вероятность того, что среди извлеченных деталей(А) ровно три окрашенных;(Б) хотя бы две окрашенных;(В) Ни одной окрашенной;(Г) Все окрашенные

Answer 1

Думаю, что ответ №2, но не уверена

Answer 2

Problem Statement

In a box, there are 15 parts, out of which 9 are colored. If 4 parts are randomly selected, we need to find the probability of the following events: (A) Exactly three colored parts are selected. (B) At least two colored parts are selected. (C) No colored parts are selected. (D) All colored parts are selected.

Solution

To solve this problem, we need to calculate the probability for each event based on the given information.

Event A: Exactly three colored parts are selected

To calculate the probability of exactly three colored parts being selected, we need to consider the number of ways we can choose 3 colored parts out of 9, multiplied by the number of ways we can choose 1 non-colored part out of 6 (since there are 15 - 9 = 6 non-colored parts). The total number of ways to choose 4 parts out of 15 is given by the binomial coefficient C(15, 4).

The probability of event A can be calculated as follows:

Probability of event A = (Number of ways to choose 3 colored parts out of 9) * (Number of ways to choose 1 non-colored part out of 6) / (Total number of ways to choose 4 parts out of 15)

Let's calculate the probability using the given information:

Number of ways to choose 3 colored parts out of 9 = C(9, 3) = 84 Number of ways to choose 1 non-colored part out of 6 = C(6, 1) = 6 Total number of ways to choose 4 parts out of 15 = C(15, 4) = 1365

Probability of event A = (84 * 6) / 1365 = 0.36842105263 Therefore, the probability of exactly three colored parts being selected is approximately 0.3684.

Event B: At least two colored parts are selected

To calculate the probability of at least two colored parts being selected, we need to consider the number of ways we can choose 2, 3, or 4 colored parts out of 9, multiplied by the number of ways we can choose the remaining non-colored parts. The total number of ways to choose 4 parts out of 15 is given by the binomial coefficient C(15, 4).

The probability of event B can be calculated as follows:

Probability of event B = (Number of ways to choose 2 colored parts out of 9) * (Number of ways to choose 2 non-colored parts out of 6) + (Number of ways to choose 3 colored parts out of 9) * (Number of ways to choose 1 non-colored part out of 6) + (Number of ways to choose 4 colored parts out of 9) / (Total number of ways to choose 4 parts out of 15)

Let's calculate the probability using the given information:

Number of ways to choose 2 colored parts out of 9 = C(9, 2) = 36 Number of ways to choose 2 non-colored parts out of 6 = C(6, 2) = 15 Number of ways to choose 3 colored parts out of 9 = C(9, 3) = 84 Number of ways to choose 1 non-colored part out of 6 = C(6, 1) = 6 Number of ways to choose 4 colored parts out of 9 = C(9, 4) = 126 Total number of ways to choose 4 parts out of 15 = C(15, 4) = 1365

Probability of event B = ((36 * 15) + (84 * 6) + 126) / 1365 = 0.87719298246 Therefore, the probability of at least two colored parts being selected is approximately 0.8772.

Event C: No colored parts are selected

To calculate the probability of no colored parts being selected, we need to consider the number of ways we can choose 4 non-colored parts out of 6. The total number of ways to choose 4 parts out of 15 is given by the binomial coefficient C(15, 4).

The probability of event C can be calculated as follows:

Probability of event C = (Number of ways to choose 4 non-colored parts out of 6) / (Total number of ways to choose 4 parts out of 15)

Let's calculate the probability using the given information:

Number of ways to choose 4 non-colored parts out of 6 = C(6, 4) = 15 Total number of ways to choose 4 parts out of 15 = C(15, 4) = 1365

Probability of event C = 15 / 1365 = 0.01098901099 Therefore, the probability of no colored parts being selected is approximately 0.0110.

Event D: All colored parts are selected

To calculate the probability of all colored parts being selected, we need to consider the number of ways we can choose 4 colored parts out of 9. The total number of ways to choose 4 parts out of 15 is given by the binomial coefficient C(15, 4).

The probability of event D can be calculated as follows:

Probability of event D = (Number of ways to choose 4 colored parts out of 9) / (Total number of ways to choose 4 parts out of 15)

Let's calculate the probability using the given information:

Number of ways to choose 4 colored parts out of 9 = C(9, 4) = 126 Total number of ways to choose 4 parts out of 15 = C(15, 4) = 1365

Probability of event D = 126 / 1365 = 0.0923076923 Therefore, the probability of all colored parts being selected is approximately 0.0923.

Summary

Based on the given information, we have calculated the following probabilities: (A) Probability of exactly three colored parts being selected = 0.3684 (B) Probability of at least two colored parts being selected = 0.8772 (C) Probability of no colored parts being selected = 0.0110 (D) Probability of all colored parts being selected = 0.0923