Question

Среди 18 машин, стоящих на парковке семь – выпуска 2013 года, остальные - более ранних годов. Случайным образом выезжают пять машин. Какова вероятность того, что среди выехавших машин три - выпуска 2013 года.

Answer 1

C=m/n = C(7,3)·C(11,2) / C(18,5), где C(n, k) = n! / [k!·(n − k)!]

С-это вероятность,м-благоприятные исходы,н-все исходы.Это общая формула по нахождению вероятностей

Факториал
Пример: 10!=10*9*8*7*6*5*4*3*2*1
Поехалииии
С=((7!/3!*(7-3)!)*(11!/2!*(11-2)!)
        .........................................=35*55/3003=0.641....=0.6
        15!/5!*(15-5)!
там цифры очееееееееень длинные поэтому не писал,но
как-то так...................

Answer 2

Problem Analysis

We have a parking lot with 18 cars, 7 of which were manufactured in 2013 and the rest were manufactured in earlier years. We randomly select 5 cars to leave the parking lot. We need to find the probability that among the 5 selected cars, 3 of them were manufactured in 2013.

Solution

To find the probability, we need to calculate the total number of possible outcomes and the number of favorable outcomes.

The total number of possible outcomes is the number of ways we can select 5 cars from the 18 cars in the parking lot. This can be calculated using the combination formula:

Total number of possible outcomes = C(18, 5)

The number of favorable outcomes is the number of ways we can select 3 cars from the 7 cars manufactured in 2013 and 2 cars from the remaining cars manufactured in earlier years. This can also be calculated using the combination formula:

Number of favorable outcomes = C(7, 3) * C(11, 2)

Finally, we can calculate the probability by dividing the number of favorable outcomes by the total number of possible outcomes:

Probability = Number of favorable outcomes / Total number of possible outcomes

Let's calculate the probability.

Calculation

The total number of possible outcomes can be calculated as follows:

Total number of possible outcomes = C(18, 5) = 18! / (5! * (18-5)!) = 856

The number of favorable outcomes can be calculated as follows:

Number of favorable outcomes = C(7, 3) * C(11, 2) = (7! / (3! * (7-3)!)) * (11! / (2! * (11-2)!)) = 35 * 55 = 1925

Therefore, the probability of selecting 3 cars manufactured in 2013 among the 5 selected cars is:

Probability = Number of favorable outcomes / Total number of possible outcomes = 1925 / 856 ≈ 0.2257

So, the probability is approximately 0.2257.

Answer

The probability that among the 5 randomly selected cars, 3 of them are manufactured in 2013 is approximately 0.2257.