Question

Из коробки, содержащей 4 синих, 3 красных и 1 зеленых карандашей, достают наугад 3 карандаша. Найти вероятность, что будут вынуты карандаши: а) одного цвета; б) разных цветов.

Answer 1

Испытание состоит в том, что из  4+3+1=8 карандашей выбирают 3 карандаша.
Это можно сделать
$C^3_8= \frac{8!}{(8-3)!\cdot 3!} = \frac{6\cdot 7\cdot 8}{6}=56$ способами
Событию А - "вынуты карандаши одного цвета" благоприятствуют  случаи:
 все три карандаша синие или красные
Это можно сделать
$C^3_4+C^3_3=4+1=5$ способами
По формуле классической вероятности
$p(A)= \frac{m}{n}= \frac{5}{56}$
n- число исходов испытания
m- число исходов благоприятствующих данному событию
Событие В-"вынуты карандаши разного цвета", один синий, один красный и один зеленый.
Число таких случаев
$C^1_4+C^1_3+C^1_1=4+3+1=8$

По формуле классической вероятности
$p(A)= \frac{m}{n}= \frac{8}{56}=\frac{1}{7}$

Answer 2

Problem Statement

Из коробки, содержащей 4 синих, 3 красных и 1 зеленый карандашей, достают наугад 3 карандаша. Найти вероятность, что будут вынуты карандаши: а) одного цвета; б) разных цветов.

Solution

To solve this problem, we need to calculate the probabilities of drawing pencils of different colors from the given box.

# Part (a): Probability of drawing pencils of the same color

To find the probability of drawing pencils of the same color, we need to consider two cases: drawing three pencils of the same color and drawing two pencils of one color and one pencil of another color.

Case 1: Drawing three pencils of the same color In this case, we can choose any of the three colors (blue, red, or green) and calculate the probability of drawing three pencils of that color.

The probability of drawing three blue pencils: - Number of ways to choose 3 blue pencils: C(4, 3) = 4 - Number of ways to choose any 3 pencils from the box: C(8, 3) = 56

The probability of drawing three red pencils: - Number of ways to choose 3 red pencils: C(3, 3) = 1 - Number of ways to choose any 3 pencils from the box: C(8, 3) = 56

The probability of drawing three green pencils: - Number of ways to choose 3 green pencils: C(1, 3) = 0 (since there is only 1 green pencil) - Number of ways to choose any 3 pencils from the box: C(8, 3) = 56

Therefore, the total probability of drawing three pencils of the same color is: P(same color) = (4 + 1 + 0) / 56 = 5 / 56 = 0.0893.

Case 2: Drawing two pencils of one color and one pencil of another color In this case, we need to consider all possible combinations of two colors and calculate the probability of drawing two pencils of one color and one pencil of another color.

The probability of drawing two blue pencils and one red pencil: - Number of ways to choose 2 blue pencils: C(4, 2) = 6 - Number of ways to choose 1 red pencil: C(3, 1) = 3 - Number of ways to choose any 3 pencils from the box: C(8, 3) = 56

The probability of drawing two blue pencils and one green pencil: - Number of ways to choose 2 blue pencils: C(4, 2) = 6 - Number of ways to choose 1 green pencil: C(1, 1) = 1 - Number of ways to choose any 3 pencils from the box: C(8, 3) = 56

The probability of drawing two red pencils and one blue pencil: - Number of ways to choose 2 red pencils: C(3, 2) = 3 - Number of ways to choose 1 blue pencil: C(4, 1) = 4 - Number of ways to choose any 3 pencils from the box: C(8, 3) = 56

The probability of drawing two red pencils and one green pencil: - Number of ways to choose 2 red pencils: C(3, 2) = 3 - Number of ways to choose 1 green pencil: C(1, 1) = 1 - Number of ways to choose any 3 pencils from the box: C(8, 3) = 56

The probability of drawing two green pencils and one blue pencil: - Number of ways to choose 2 green pencils: C(1, 2) = 0 (since there is only 1 green pencil) - Number of ways to choose 1 blue pencil: C(4, 1) = 4 - Number of ways to choose any 3 pencils from the box: C(8, 3) = 56

The probability of drawing two green pencils and one red pencil: - Number of ways to choose 2 green pencils: C(1, 2) = 0 (since there is only 1 green pencil) - Number of ways to choose 1 red pencil: C(3, 1) = 3 - Number of ways to choose any 3 pencils from the box: C(8, 3) = 56

Therefore, the total probability of drawing two pencils of one color and one pencil of another color is: P(two colors) = (6 + 6 + 3 + 3 + 4 + 3) / 56 = 25 / 56 = 0.4464.

# Part (b): Probability of drawing pencils of different colors

To find the probability of drawing pencils of different colors, we need to consider all possible combinations of three different colors and calculate the probability of drawing one pencil of each color.

The probability of drawing one blue, one red, and one green pencil: - Number of ways to choose 1 blue pencil: C(4, 1) = 4 - Number of ways to choose 1 red pencil: C(3, 1) = 3 - Number of ways to choose 1 green pencil: C(1, 1) = 1 - Number of ways to choose any 3 pencils from the box: C(8, 3) = 56

Therefore, the probability of drawing pencils of different colors is: P(different colors) = (4 * 3 * 1) / 56 = 12 / 56 = 0.2143.

Summary

The probabilities of drawing pencils from the given box are as follows: a) The probability of drawing pencils of the same color is approximately 0.0893. b) The probability of drawing pencils of different colors is approximately 0.2143.

Please note that these probabilities are approximate and may vary slightly due to rounding.