Question

Помогите пожалуйста решить из трамвайного парка в случайном порядке последовательно выходят 3 трамвая маршрута номер 1 и 7 трамваев маршрута номер 2. найти вероятность того что вторым по порядку выходит на линию трамвай маршрута номер 1

Answer 1

После того, как выйдет трамвай под №1, то будет 6 маршрутов. Значит, вероятность находится отношением благ.соб ко всем: 
1/6=1.2

Answer 2

Problem Analysis

We are given that 3 trams of route number 1 and 7 trams of route number 2 are leaving the tram park in random order. We need to find the probability that the second tram to leave is from route number 1.

Solution

To solve this problem, we can use the concept of permutations. The total number of ways the trams can leave the park is the total number of permutations of the 10 trams (3 from route 1 and 7 from route 2). The number of ways the second tram can be from route 1 is the number of permutations of 2 trams from route 1 and 7 trams from route 2.

Let's calculate the probabilities step by step:

1. Calculate the total number of permutations of 10 trams: - The total number of trams is 10 (3 from route 1 and 7#### Calculating the Probability of a Tram Route

To calculate the probability of the second tram to come out being from route 1, we can use the concept of permutations.

Permutations are the number of ways to arrange a set of items in a specific order. In this case, we want to find the probability of the second tram being from route 1 out of all possible arrangements.

The total number of ways the trams can come out in random order is the total number of trams factorial, which is 6 (3 trams from route 1 and 3 trams from route 2).

The number of favorable outcomes (the second tram being from route 1) is the number of ways to choose 1 tram from route 1 out of 3, multiplied by the number of ways to choose 1 tram from route 2 out of 2.

Calculation

The probability can be calculated using the formula:

Probability = (Number of favorable outcomes) / (Total number of outcomes)

Total number of outcomes = 6! = 720

Number of favorable outcomes = (3! * 2!) = 12

Probability = 12 / 720 = 1 / 60

Therefore, the probability of the second tram being from route 1 is 1/60.