
Среди 11 лотерейных билетов 8 выигрышных. На удачу взяли 4 билета. Определить вероятность того, что
среди них 3 выигрышных.

Ответы на вопрос




Problem Analysis
We are given that out of 11 lottery tickets, 8 are winning tickets. We also know that 4 tickets are chosen randomly. We need to determine the probability that among these 4 chosen tickets, 3 are winning tickets.Solution
To solve this problem, we can use the concept of combinations. The number of ways to choose 3 winning tickets out of 8 is given by the combination formula:C(n, k) = n! / (k! * (n-k)!)
Where n is the total number of winning tickets (8) and k is the number of winning tickets chosen (3).
Similarly, the number of ways to choose 1 losing ticket out of 3 is given by:
C(n, k) = n! / (k! * (n-k)!)
Where n is the total number of losing tickets (3) and k is the number of losing tickets chosen (1).
The total number of ways to choose 4 tickets out of 11 is given by:
C(n, k) = n! / (k! * (n-k)!)
Where n is the total number of tickets (11) and k is the number of tickets chosen (4).
To find the probability, we divide the number of favorable outcomes (choosing 3 winning tickets and 1 losing ticket) by the total number of possible outcomes (choosing any 4 tickets).
Let's calculate the probability.
Calculation
The number of ways to choose 3 winning tickets out of 8 is:C(8, 3) = 8! / (3! * (8-3)!) = 56
The number of ways to choose 1 losing ticket out of 3 is:
C(3, 1) = 3! / (1! * (3-1)!) = 3
The total number of ways to choose 4 tickets out of 11 is:
C(11, 4) = 11! / (4! * (11-4)!) = 330
Therefore, the probability of choosing 3 winning tickets and 1 losing ticket out of 4 is:
P = (number of favorable outcomes) / (total number of possible outcomes) = (56 * 3) / 330 = 168 / 330
Simplifying the fraction, we get:
P = 0.5091
So, the probability of choosing 3 winning tickets and 1 losing ticket out of 4 is approximately 0.5091.
Answer
The probability of choosing 3 winning tickets and 1 losing ticket out of 4 is approximately 0.5091.

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