Question

Среди 9 лотерейных билетов 4 выигрышных. Наудачу взяли 4 билета. Определить вероятность того, что среди них 3 выигрышных.

Answer 1

$P= \frac{C_4^3*C_4^1}{C_9^4}=\frac{\frac{4!}{(4-3)!3!} * \frac{4!}{(4-1)!1!}}{ \frac{9!}{(9-4)!*4!}}=\frac{\frac{4!}{1!3!}*\frac{4!}{3!1!}}{\frac{9!}{5!*4!}}=\frac{4*4}{\frac{9*8*7*6}{4!}}=$

$=\frac{4*4*4!}{9*8*7*6}=\frac{4*4!}{9*2*7*2*3}=\frac{4!}{9*7*3}=\frac{4*3*2*1}{9*7*3}=\frac{4*2}{9*7}=\frac{8}{63}$

Answer 2

Problem Analysis

We are given that there are 9 lottery tickets, out of which 4 are winning tickets. We randomly select 4 tickets and we need to determine the probability that 3 of them are winning tickets.

Solution

To solve this problem, we can use the concept of combinations. The probability of selecting 3 winning tickets out of 4 is equal to the number of ways to select 3 winning tickets divided by the total number of ways to select 4 tickets.

The number of ways to select 3 winning tickets out of 4 is given by the combination formula: C(n, k) = n! / (k!(n-k)!), where n is the total number of items and k is the number of items to be selected.

In this case, we have 4 winning tickets and we want to select 3 of them. Therefore, the number of ways to select 3 winning tickets out of 4 is: C(4, 3) = 4! / (3!(4-3)!) = 4.

The total number of ways to select 4 tickets out of 9 is given by the combination formula: C(n, k) = n! / (k!(n-k)!), where n is the total number of items and k is the number of items to be selected.

In this case, we have 9 tickets and we want to select 4 of them. Therefore, the total number of ways to select 4 tickets out of 9 is: C(9, 4) = 9! / (4!(9-4)!) = 126.

Now, we can calculate the probability by dividing the number of ways to select 3 winning tickets by the total number of ways to select 4 tickets: P = 4 / 126.

Calculation

Let's calculate the probability:

P = 4 / 126

P ≈ 0.0317

Answer

The probability that among the 4 randomly selected tickets, 3 of them are winning tickets is approximately 0.0317.

Please note that the above calculation assumes that each ticket has an equal chance of being selected and that the selection is done without replacement (i.e., once a ticket is selected, it is not put back into the pool of tickets).