The urn contains 9 numbered balls. Take 4 balls at random. Find the probability that among

The urn contains 9 numbered balls. Take 4 balls at random. Find the probability that among the taken balls 3 will have even numbers?

There are 9 balls in the urn in total, so 5 of them have odd numbers and 4 are even. We choose 4 balls and we are interested in the event in which 1 ball came from the first group and 3 balls from the second group.

If we consider a group of selected balls (3 even and 1 odd), and consider how many permutations can be among them, then it is obvious that there will be only 4 options. Because an odd ball can be the first, second, third or fourth. This means that the created group can be obtained in four possible ways by pulling out the balls one by one.

Now let’s consider the probability with which we would draw an odd ball for the first time (5 chances out of 9), and 3 subsequent even balls (4 chances out of 8, 3 chances out of 7, 2 chances out of 6).

P (A) = (5/9) * (4/8) * (3/7) * (2/6) = 5 / (2 * 9 * 7) = 5/126

P = 4 * P (A) = 10/63 = 0.15873

Answer: the probability that 3 balls will have even numbers is 15.87%