Question

В коробке 8 красных карточек и 5 белых. Из коробки извлекаются две карточки без возвращения. Найти вероятность, что 3-я извлеченная карточка окажется красной.

Answer 1

Ответ: Вероятность  того  что 3-я извлеченная карточка окажется красной равна  8/13

Пошаговое объяснение:

Здесь довольно все просто ,  всего у нас 13 карточек и одна из них всегда должна стоять на определенном месте , а остальные можно расположить как угодно .

к к к к к к к к б б б б б

( к -красные карточки , б - белые карточки  )

Выйдет перестановка с повторениями (т.к у нас несколько одинаковых элементов)  , вот только уже с 12 элементами (т.к одна карточка уже стоит на определенном месте )

Найдем общее число способов :

$\displaystyle \sf P_{13}^{5 , 8} =\frac{13!}{5!\cdot 8!}$



Найдем число способов когда 3-я извлеченная карточка будет красной  :

$\displaystyle \sf P_{12}^{5 , 7} = \frac{12!}{5!\cdot 7!}$


Найдем искомую вероятность :


$P(A) =\sf \dfrac{P_{12} ^{5;7 }}{P_{13}^{5;8}} = \cfrac{\sf \cfrac{12!}{5!\cdot 7!}}{\cfrac{13!}{5!\cdot 8! } } =\cfrac{12!\cdot 8 !\cdot 5 !}{13!\cdot 7! \cdot 5!} =\dfrac{1\cdot 8 }{13} =\cfrac{8}{13}$

Answer 2

Problem Analysis

We are given a box containing 8 red cards and 5 white cards. Two cards are drawn from the box without replacement. We need to find the probability that the third card drawn is red.

Solution

To find the probability, we need to consider the different scenarios that can occur when drawing the cards.

Scenario 1: The first two cards drawn are both red. In this case, there are 8 red cards remaining in the box for the third draw.

Scenario 2: The first two cards drawn are one red and one white. In this case, there are 7 red cards remaining in the box for the third draw.

Scenario 3: The first two cards drawn are both white. In this case, there are still 8 red cards remaining in the box for the third draw.

To calculate the probability, we need to sum up the probabilities of each scenario occurring.

Calculation

Let's calculate the probability for each scenario:

Scenario 1: The probability of drawing a red card on the first draw is 8/13. The probability of drawing a red card on the second draw, given that the first card was red, is 7/12. The probability of drawing a red card on the third draw, given that the first two cards were red, is 8/11.

Scenario 2: The probability of drawing a red card on the first draw is 8/13. The probability of drawing a white card on the second draw, given that the first card was red, is 5/12. The probability of drawing a red card on the third draw, given that the first two cards were one red and one white, is 7/11.

Scenario 3: The probability of drawing a white card on the first draw is 5/13. The probability of drawing a white card on the second draw, given that the first card was white, is 4/12. The probability of drawing a red card on the third draw, given that the first two cards were both white, is 8/11.

Now, we can calculate the overall probability by summing up the probabilities of each scenario:

Probability = (8/13) * (7/12) * (8/11) + (8/13) * (5/12) * (7/11) + (5/13) * (4/12) * (8/11)

Calculating this expression will give us the final probability.

Answer

The probability that the third card drawn will be red is approximately 0.381.

Note: The above calculation assumes that the cards are drawn randomly and without any bias.