Question

В коробке 10 зеленых и 4 желтых яблока. Наудачу вынимают 5 яблок. Составить закон распределения, математическое ожидание и дисперсию числа зеленых яблок среди вынутых. Найти вероятность, что среди вынутых яблок будет более двух зеленых.

Answer 1

10+4-5=9
Ответ девять яблок

Answer 2

Problem Statement

In a box, there are 10 green apples and 4 yellow apples. If 5 apples are randomly selected from the box, we need to determine the probability distribution, the expected value, and the variance of the number of green apples among the selected ones. We also need to find the probability that there will be more than two green apples among the selected ones.

Probability Distribution

To determine the probability distribution, we need to calculate the probabilities for each possible outcome. Let's denote the number of green apples among the selected ones as X.

The possible values for X are 0, 1, 2, 3, 4, and 5. We can calculate the probabilities using the hypergeometric distribution formula:

P(X = k) = (C(10, k) * C(4, 5 - k)) / C(14, 5)

where C(n, r) represents the number of combinations of n items taken r at a time.

Let's calculate the probabilities for each value of X:

- P(X = 0) = (C(10, 0) * C(4, 5 - 0)) / C(14, 5) - P(X = 1) = (C(10, 1) * C(4, 5 - 1)) / C(14, 5) - P(X = 2) = (C(10, 2) * C(4, 5 - 2)) / C(14, 5) - P(X = 3) = (C(10, 3) * C(4, 5 - 3)) / C(14, 5) - P(X = 4) = (C(10, 4) * C(4, 5 - 4)) / C(14, 5) - P(X = 5) = (C(10, 5) * C(4, 5 - 5)) / C(14, 5)

Let's calculate these probabilities:

- P(X = 0) = (C(10, 0) * C(4, 5 - 0)) / C(14, 5) = (1 * 1) / 2002 = 1 / 2002 - P(X = 1) = (C(10, 1) * C(4, 5 - 1)) / C(14, 5) = (10 * 4) / 2002 = 40 / 2002 - P(X = 2) = (C(10, 2) * C(4, 5 - 2)) / C(14, 5) = (45 * 6) / 2002 = 270 / 2002 - P(X = 3) = (C(10, 3) * C(4, 5 - 3)) / C(14, 5) = (120 * 4) / 2002 = 480 / 2002 - P(X = 4) = (C(10, 4) * C(4, 5 - 4)) / C(14, 5) = (210 * 1) / 2002 = 210 / 2002 - P(X = 5) = (C(10, 5) * C(4, 5 - 5)) / C(14, 5) = (252 * 1) / 2002 = 252 / 2002

Therefore, the probability distribution is as follows:

- P(X = 0) = 1 / 2002 - P(X = 1) = 40 / 2002 - P(X = 2) = 270 / 2002 - P(X = 3) = 480 / 2002 - P(X = 4) = 210 / 2002 - P(X = 5) = 252 / 2002

Mathematical Expectation (Expected Value)

The expected value, denoted as E(X), represents the average value of X. It is calculated by multiplying each possible value of X by its corresponding probability and summing them up.

E(X) = Σ(X * P(X))

Let's calculate the expected value:

- E(X) = (0 * P(X = 0)) + (1 * P(X = 1)) + (2 * P(X = 2)) + (3 * P(X = 3)) + (4 * P(X = 4)) + (5 * P(X = 5)) - E(X) = (0 * (1 / 2002)) + (1 * (40 / 2002)) + (2 * (270 / 2002)) + (3 * (480 / 2002)) + (4 * (210 / 2002)) + (5 * (252 / 2002)) - E(X) = 3.57 (rounded to two decimal places)

Therefore, the expected value of the number of green apples among the selected ones is approximately 3.57.

Variance

The variance, denoted as Var(X), measures the spread or dispersion of the values of X around the expected value. It is calculated by subtracting the square of the expected value from the expected value of the square of X.

Var(X) = E(X^2) - (E(X))^2

To calculate the variance, we need to calculate E(X^2) first. E(X^2) is calculated by multiplying each possible value of X squared by its corresponding probability and summing them up.

E(X^2) = Σ(X^2 * P(X))

Let's calculate E(X^2):

- E(X^2) = (0^2 * P(X = 0)) + (1^2 * P(X = 1)) + (2^2 * P(X = 2)) + (3^2 * P(X = 3)) + (4^2 * P(X = 4)) + (5^2 * P(X = 5)) - E(X^2) = (0^2 * (1 / 2002)) + (1^2 * (40 / 2002)) + (2^2 * (270 / 2002)) + (3^2 * (480 / 2002)) + (4^2 * (210 / 2002)) + (5^2 * (252 / 2002)) - E(X^2) = 11.43 (rounded to two decimal places)

Now, let's calculate the variance:

- Var(X) = E(X^2) - (E(X))^2 - Var(X) = 11.43 - (3.57)^2 - Var(X) = 11.43 - 12.7449 - Var(X) = -1.3149 (rounded to four decimal places)

The variance of the number of green apples among the selected ones is approximately -1.3149.

Probability of More Than Two Green Apples

To find the probability that there will be more than two green apples among the selected ones, we need to calculate the sum of the probabilities for X = 3, X = 4, and X = 5.

Let's calculate this probability:

- P(X > 2) = P(X = 3) + P(X = 4) + P(X = 5) - P(X > 2) = (480 / 2002) + (210 / 2002) + (252 / 2002) - P(X > 2) = 0.4705 (rounded to four decimal places)

Therefore, the probability that there will be more than two green apples among the selected ones is approximately 0.4705.

Conclusion

In summary, the probability distribution of the number of green apples among the selected ones is as follows: - P(X = 0) = 1 / 2002 - P(X = 1) = 40 / 2002 - P(X = 2) = 270 / 2002 - P(X = 3) = 480 / 2002 - P(X = 4) = 210 / 2002 - P(X = 5) = 252 / 2002

The expected value of the number of green apples among the selected ones is approximately 3.57, and the variance is approximately -1.3149. The probability that there will be more than two green apples among the selected ones is approximately 0.4705.