Question

Незадачливый курильщик, чтобы не мусорить на вечеринке, складывал сожженные спички обратно в коробок. В результате там оказалось 6 хороших и 2 сожженных спички. Выйдя на улицу, он решил прикурить, случайным образом выбирая спички из коробка и выбрасывая сожженные спички при необходимости. Найти вероятность того, что только третья извлеченная из коробка спичка окажется не сожженной.

Answer 1

Вот и ответ вероятность равна (4/10)*(3/9)*(6/8).

Answer 2

This is a question about probability theory. I searched the web for similar questions and found some answers that might help you. Here is a summary of what I found:

The problem can be solved by using the rule of product and the conditional probability. The rule of product states that the probability of two independent events happening together is equal to the product of their individual probabilities. The conditional probability states that the probability of an event happening given that another event has already happened is equal to the ratio of the joint probability of both events and the probability of the given event.

In this case, we want to find the probability that only the third match drawn from the box is good, which means that the first two matches must be burnt. Let A be the event that the first match is burnt, B be the event that the second match is burnt, and C be the event that the third match is good. Then we can write:

P(A and B and C) = P(A) * P(B|A) * P(C|A and B)

where P(A and B and C) is the joint probability of all three events, P(A) is the probability of drawing a burnt match first, P(B|A) is the conditional probability of drawing a burnt match second given that a burnt match was drawn first, and P(C|A and B) is the conditional probability of drawing a good match third given that two burnt matches were drawn before.

To calculate these probabilities, we need to know how many matches are in the box at each stage. Initially, there are 8 matches in total, 2 of which are burnt and 6 of which are good. Therefore, P(A) = 2/8 = 1/4. After drawing a burnt match, there are 7 matches left in the box, 1 of which is burnt and 6 of which are good. Therefore, P(B|A) = 1/7. After drawing another burnt match, there are 6 matches left in the box, all of which are good. Therefore, P(C|A and B) = 6/6 = 1.

Plugging these values into the formula, we get:

P(A and B and C) = (1/4) * (1/7) * (1) = 1/28

Therefore, the probability that only the third match drawn from the box is good is 1/28.

You can find more details and explanations on these websites: [Uchi.ru](https://uchi.ru/otvety/questions/nezadachliviy-kurilschik-chtobi-ne-musorit-na-vecherinke-skladival-sozhzhennie-spichki-ob), [Mozgovnet.com](https://mozgovnet.com/task/10537739), [Obrazovalka.com](https://obrazovalka.com/qa/algebra/274094-nezadachlivyj-kurilshhik-chtob-ne-musorit-na-vecherinke-skladyval-sozhzhennye-spichki-nazad.html). I hope this helps you understand the problem better.