Question

В экзаменационные билеты включено 20 экзаменационных вопросов, из которых студент знает: 14. Экзаменационный билет содержит 3 вопроса. Какова вероятность того, что студент ответит хотя бы на один вопрос?

Answer 1

X____________20
3______________10

10x=60
x=6%

Answer 2

The question you asked is a probability problem that can be solved using the formula of combinations. I found a similar problem on Google Colab, where the solution is explained in detail. Here is a summary of the steps:

- First, we need to calculate the total number of possible tickets, which is the number of ways to choose 3 questions out of 20. This is given by the formula C(20, 3) = 20! / (3! * 17!) = 1140. - Next, we need to calculate the number of favorable tickets, which are the tickets that contain at least one question that the student knows. There are two ways to do this: either by adding the number of tickets that contain 1, 2, or 3 questions that the student knows, or by subtracting the number of tickets that contain only questions that the student does not know from the total number of tickets. The second way is easier, since there are fewer cases to consider. The number of tickets that contain only questions that the student does not know is the number of ways to choose 3 questions out of 6 that the student does not know. This is given by the formula C(6, 3) = 6! / (3! * 3!) = 20. Therefore, the number of favorable tickets is 1140 - 20 = 1120. - Finally, we need to divide the number of favorable tickets by the total number of tickets to get the probability of the event. This is given by the formula P = 1120 / 1140 = 0.9825. This means that the student has a 98.25% chance of answering at least one question in the ticket.