Question

Вероятность того, что студент Вагонов сдаст экзамен по теории вероятностей - 0,6, студент Рельсов - 0,2, студентка Шпалова -0,4. Какова вероятность того, что экзамен сдадут хотя бы двое из них? Пожалуйста подробно

Answer 1

Суммарная вероятность равна сумме
что сдадут все трое
что сдадут первые два не сдаст третий
что сдадут 1 и3 не сдаст второй
что сдадут 2 и 3 не сдаст первый.
Р=0.6*0.2*0.4+
0.6*0.2*(1-0.4)+
0.6*(1-0.2)*0.4+
(1-0.6)*0.2*0.4=
0.344
надеюсь в арифметике не ошибся ;)

Answer 2

Calculation of the Probability

To calculate the probability that at least two of the three students will pass the exam, we need to consider the probabilities of different combinations of students passing the exam. Let's denote the students as follows: - Student Vagonov: P(V) = 0.6 - Student Relsov: P(R) = 0.2 - Student Shpalova: P(S) = 0.4

We need to calculate the probability that at least two of them will pass the exam.

Probability of Two Students Passing

To calculate the probability of two students passing the exam, we need to consider all possible combinations of two students out of the three. There are three possible combinations: Vagonov and Relsov (VR), Vagonov and Shpalova (VS), and Relsov and Shpalova (RS).

The probability of Vagonov and Relsov passing the exam is given by: P(VR) = P(V) * P(R) = 0.6 * 0.2 = 0.12

The probability of Vagonov and Shpalova passing the exam is given by: P(VS) = P(V) * P(S) = 0.6 * 0.4 = 0.24

The probability of Relsov and Shpalova passing the exam is given by: P(RS) = P(R) * P(S) = 0.2 * 0.4 = 0.08

Probability of All Three Students Passing

To calculate the probability of all three students passing the exam, we need to consider the intersection of the probabilities of each student passing the exam.

The probability of all three students passing the exam is given by: P(VRS) = P(V) * P(R) * P(S) = 0.6 * 0.2 * 0.4 = 0.048

Probability of at Least Two Students Passing

To calculate the probability of at least two students passing the exam, we need to consider the probabilities of two students passing (VR, VS, RS) and the probability of all three students passing (VRS).

The probability of at least two students passing the exam is given by the sum of these probabilities: P(at least two students passing) = P(VR) + P(VS) + P(RS) + P(VRS) = 0.12 + 0.24 + 0.08 + 0.048 = 0.488

Therefore, the probability that at least two of the three students will pass the exam is 0.488 or 48.8%.

[[1]]