Question

В коробке лежат 6 белых и 7 чёрных шаров. Наугад вынимают 2 шара. Найти вероятность события: по крайней мере один шар белый

Answer 1

Р=m/n
M=6благоприятные события
N=13все возможные
P=6/13

Answer 2

Problem Statement

В коробке лежат 6 белых и 7 черных шаров. Наугад вынимают 2 шара. Найти вероятность события: по крайней мере один шар белый.

Solution

To find the probability of at least one white ball being drawn, we can calculate the probability of the complementary event, which is the probability of drawing no white balls.

Let's calculate the probability of drawing no white balls in two draws.

The total number of balls in the box is 6 white balls + 7 black balls = 13 balls.

The probability of drawing a black ball on the first draw is 7 black balls / 13 total balls.

After the first draw, there are 12 balls left in the box, with 6 white balls and 6 black balls.

The probability of drawing a black ball on the second draw, given that the first draw was black, is 6 black balls / 12 remaining balls.

To find the probability of drawing no white balls in two draws, we multiply the probabilities of the individual draws:

P(no white balls) = (7/13) * (6/12) = 42/156 = 7/26

The probability of at least one white ball being drawn is equal to 1 minus the probability of drawing no white balls:

P(at least one white ball) = 1 - P(no white balls) = 1 - 7/26 = 19/26

Therefore, the probability of drawing at least one white ball is 19/26.

Answer: The probability of drawing at least one white ball is 19/26.