Question

ПОМОГИТЕ ПОЖАЛУЙСТА Было два ящика в 1 ящике было 3 белых и 5 черных шаров, во 2 ящике было 4 белых и 3 черных шаров. Из 1ого ящика достали 2 шара из второго один . 1)Какова вероятность того что все шарики которые вытащили белые? 2)Какова вероятность тог что 1 шарик черный? Бросают 3 игральных кубика . Какова вероятность что 3 выпадет более одного раза?

Answer 1

1)
8/3+7/4= 32/12+21/12=53/12=4+5/12
2) Соответственно половина
2+5/24

Answer 2

Probability of Drawing All White Balls from the First Box

To calculate the probability of drawing all white balls from the first box, we can use the formula for conditional probability. Let's denote the events as follows: - A: Drawing a white ball from the first box on the first draw - B: Drawing a white ball from the first box on the second draw, given that the first draw was a white ball

The probability of event A is the number of white balls in the first box divided by the total number of balls in the first box. The probability of event B is the number of white balls remaining in the first box after the first draw, divided by the total number of balls remaining in the first box.

The probability of both events A and B occurring is the product of the individual probabilities.

Calculation: - Probability of drawing a white ball from the first box on the first draw: 3 white balls out of 8 total balls - Probability of drawing a white ball from the first box on the second draw, given the first draw was a white ball: 2 white balls out of 7 remaining balls

The probability of drawing all white balls from the first box is the product of these probabilities.

Probability of drawing all white balls from the first box: - \( P(A) = \frac{3}{8} \) - \( P(B|A) = \frac{2}{7} \) - \( P(A \cap B) = P(A) \times P(B|A) = \frac{3}{8} \times \frac{2}{7} = \frac{3}{28} \)

Probability of Drawing At Least One Black Ball from the First Box

To calculate the probability of drawing at least one black ball from the first box, we can use the complement rule. The complement of drawing all white balls from the first box is drawing at least one black ball from the first box.

Calculation: The probability of drawing at least one black ball from the first box is 1 minus the probability of drawing all white balls from the first box.

Probability of drawing at least one black ball from the first box: - \( P(\text{at least one black ball from the first box}) = 1 - P(\text{all white balls from the first box}) = 1 - \frac{3}{28} = \frac{25}{28} \)

Probability of Rolling More Than One Three with Three Dice

To calculate the probability of rolling more than one three with three dice, we can use the complement rule. The complement of rolling more than one three is rolling one or fewer threes.

Calculation: The probability of rolling one or fewer threes with three dice can be calculated by finding the probability of rolling no threes and the probability of rolling exactly one three, and then adding these probabilities together.

Probability of rolling one or fewer threes with three dice: - Probability of rolling no threes: \( \left(\frac{5}{6}\right)^3 \) - Probability of rolling exactly one three: \( 3 \times \left(\frac{1}{6}\right) \times \left(\frac{5}{6}\right)^2 \)

The probability of rolling more than one three with three dice is 1 minus the probability of rolling one or fewer threes.

Probability of rolling more than one three with three dice: - \( P(\text{more than one three with three dice}) = 1 - \left(\left(\frac{5}{6}\right)^3 + 3 \times \left(\frac{1}{6}\right) \times \left(\frac{5}{6}\right)^2\right) \)

I hope this helps! If you have any further questions, feel free to ask.