Question

Студент знает 60 вариантов из 80. Билет содержит 2 вопроса. Найти варианты того ,что 1 )студент будет знать хотя бы один вопрос экзаменационного билета 2) что экзамен будет сдан ,если для этого достаточно ответить на оба вопроса своего билета или на один вопрос из своего билета и на 1 вопрос из дополнительного билета

Answer 1

1) Эта вероятность равна 1 - вероятность того, что он не знает оба вопроса билета. Посчитаем эту вероятность.

80 - 60 = 20 - вопросов он не знает.

20/80 - вероятность того, что он не знает первый вопрос.

(20 - 1)/(80 - 1) - вероятность, что не знает второй при том, что не знает первый.

(20/80) * (19/79) - вероятность того, что  не знает билет.

 = 19/316 ~ 0.06  = 6 процентов.

100 - 6 = 94 процентов, что знает хотя бы один вопрос.

2. В своем билете таким образом должны быть оба известных вопроса.

(60/80) * (59/79)

Или один из своего - (60/80) * (20/79) + (20/80) * (60/79)
Один из дополнительного -
- (59/78) * (19/77) + (19/78) * (59/77)

Итого, искомая вероятность есть:

(60/80) * (59/79)  + ((60/80) * (20/79) + (20/80) * (60/79)) *( (59/78) * (19/77) + (19/78) * (59/77)) ~ 0.70 = 70 процентов 

Answer 2

Finding Variants for the Student's Exam Scenario

To find the variants for the given scenario, we can use combinatorics to calculate the number of ways the student can answer the questions on the exam ticket.

1. Student Knowing At Least One Question on the Exam Ticket: - To find the number of ways the student can know at least one question on the exam ticket, we can use the principle of inclusion-exclusion. - The total number of ways the student can answer the questions on the exam ticket is given by 80 choose 60, denoted as C(80, 60). - The number of ways the student can answer both questions incorrectly is given by 20 choose 2, denoted as C(20, 2). - Therefore, the number of ways the student can know at least one question on the exam ticket is C(80, 60) - C(20, 2).

Calculation: - C(80, 60) = 80! / (60! * (80-60)!) - C(20, 2) = 20! / (2! * (20-2)!) - Number of ways the student can know at least one question = C(80, 60) - C(20, 2)

2. Passing the Exam by Answering Both Questions on the Exam Ticket or One Question on the Exam Ticket and One Question on an Additional Ticket: - To find the number of ways the student can pass the exam, we need to consider the scenarios where the student answers both questions on the exam ticket, as well as the scenarios where the student answers one question on the exam ticket and one question on an additional ticket. - We can calculate the total number of ways the student can pass the exam by summing the number of ways for each scenario.

Calculation: - Number of ways to pass by answering both questions on the exam ticket = C(60, 2) - Number of ways to pass by answering one question on the exam ticket and one question on an additional ticket = C(60, 1) * C(20, 1) - Total number of ways to pass the exam = Number of ways to pass by answering both questions + Number of ways to pass by answering one question on each ticket

Conclusion

The number of ways the student can know at least one question on the exam ticket and the number of ways the student can pass the exam by answering both questions on the exam ticket or one question on the exam ticket and one question on an additional ticket can be calculated using combinatorics.