Question

1. Щосуботи в переході залізничного вокзалу грають безробітні музиканти перший 3 Імовірністю 0,7, другий з імовірністю 0,4. Яка ймовірність того, що в найближчу суботу в переході а) будуть грати обоє музикантів, б) буде грати хоча б один з них, в) буде грати тільки перший музикант, 2. Студент Сидоренко виришує задачу без помилок з імовірністю 0,9, а Антоненко - з Імовірністю 0,6. На контрольній роботі було запропоновано три задачі. Яка ймовірність, що Антоненко виршив без помилок більше задач, ніж Сидоренко?

Answer 1

Probability of Musicians Playing in the Railway Station Passage

To calculate the probability of both musicians playing, either one of them playing, or only the first musician playing, we need to consider the probabilities of each musician playing individually.

Let's denote the first musician as M1 and the second musician as M2.

The probability of M1 playing is 0.7, and the probability of M2 playing is 0.4.

a) To calculate the probability that both musicians will play, we multiply their individual probabilities:

P(both musicians playing) = P(M1 playing) * P(M2 playing) = 0.7 * 0.4 = 0.28

b) To calculate the probability that at least one musician will play, we can use the complement rule. The complement of "at least one musician playing" is "neither musician playing." So we subtract the probability of neither musician playing from 1:

P(at least one musician playing) = 1 - P(neither musician playing)

To calculate the probability of neither musician playing, we multiply the probabilities of both musicians not playing:

P(neither musician playing) = (1 - P(M1 playing)) * (1 - P(M2 playing)) = (1 - 0.7) * (1 - 0.4) = 0.3 * 0.6 = 0.18

Therefore,

P(at least one musician playing) = 1 - 0.18 = 0.82

c) To calculate the probability that only the first musician will play, we need to consider the probability of the first musician playing and the second musician not playing:

P(only M1 playing) = P(M1 playing) * (1 - P(M2 playing)) = 0.7 * (1 - 0.4) = 0.7 * 0.6 = 0.42

Probability of Antonenko Solving More Problems Correctly Than Sidorov

To calculate the probability that Antonenko solves more problems correctly than Sidorov, we need to consider their individual probabilities of solving each problem correctly.

Let's denote the probability of Antonenko solving a problem correctly as P(A) and the probability of Sidorov solving a problem correctly as P(S).

The probability of Antonenko solving a problem correctly is 0.6, and the probability of Sidorov solving a problem correctly is 0.9.

To calculate the probability that Antonenko solves more problems correctly than Sidorov, we can use the binomial distribution.

Since there are three problems, we need to consider all possible outcomes where Antonenko solves more problems correctly than Sidorov: 2 out of 3 and 3 out of 3.

The probability of Antonenko solving 2 out of 3 problems correctly is calculated as follows:

P(Antonenko solves 2 out of 3 correctly) = P(A)^2 * (1 - P(A)) + 2 * P(A)^2 * (1 - P(A)) = 0.6^2 * (1 - 0.6) + 2 * 0.6^2 * (1 - 0.6) = 0.216

The probability of Antonenko solving all 3 problems correctly is:

P(Antonenko solves all 3 correctly) = P(A)^3 = 0.6^3 = 0.216

Therefore, the total probability that Antonenko solves more problems correctly than Sidorov is the sum of these two probabilities:

P(Antonenko solves more problems correctly than Sidorov) = 0.216 + 0.216 = 0.432

So, the probability that Antonenko solves more problems correctly than Sidorov is 0.432.