The difference between taking three balls from the pocket and not putting them back to take three ba

There is a difference in probability and outcome between taking three balls from the bag and not putting them back in a row.

1.Probability: Assuming that the number of balls in the bag is 3, when three balls are taken in a row without putting them back, the probability of the first ball being taken out is 1 3, the probability of the second ball being taken out is 2 3, and the probability of the last ball being taken out is 100%.

In the case of taking any three balls from the bag, the probability of each ball is 1 3.

2.Result: If there is a requirement for the order (e.g. three red balls are required to be taken out) without putting back the three balls, it is different from taking any three balls from the bag.

In this case, the results may be different between the two ways of taking the ball. However, if there is no order requirement and only the final three balls are taken, then the result will be the same in both ways.

In general, there is a difference in probability and outcome between taking three balls from the pocket and not putting three balls in a row, and whether there is a requirement for order will affect the similarity of the two.

To give you a specific example, if there are 3 balls in the bag, the probability of you taking any one is 1 3, and if you don't put it back, the probability of going the first time is the same as 1 3, the second time is 1 2, and the last time is 100%.

Hope it helps!

Taking any three balls is similar to treating three as a whole.

If you don't put back the three in a row, the probability between the consecutive ones will be affected.

This needs to be analyzed on a case-by-case basis.

Summary. Hello, it's an honor for my entertainment expert Ah Xuan to answer your questions, it takes a little time to sort out the answers, and you will be replied within a few minutes, please be patient

There are three balls in a bag, marked with the numbers 1,2,2 in turn, take one from any of them, do not let go of the service back to the bag, take one again, let each time you take the ball, the probability of each ball being taken to the wild regret quarrel is equal, with x, y respectively record the number marked on the ball on the first and second time, and find the distribution law and distribution function of ( , y).

Hello, it's a great honor for me to entertain Ah Xuan to answer questions for you, it takes a little time to sort out the answer stool to talk about the case, and you will be replied within a few minutes, please wait patiently

Hello dear, explained in the figure below.

Summary. There are three balls in a bag, marked with the numbers 1, 2, 2 in turn, take one of them, do not put them back into the bag, take one more ball, and set each ball to take each ball.

There are three balls in a bag, marked with the number 1,2,2 in turn, take one from any of them, do not put Sun Qiao back into the bag, and then take one, set each time you take the ball, each ball is so dear, I hope that it can help you to be acre.

If you take the number of red balls x, then x can take the value of 1, 2, 3, 4. Because there are only three whites, the fourth time it will definitely be red.

p（x=1）=7/10，p（x=2）=3/10*7/9=7/30.

p（x=3）=3/10*2/9*7/8=7/120.

p（x=4）=3/10*2/9*1/8=1/120.

I got his distribution column.

Experience is correct.

For reference, please smile.

p(ab) means the probability of getting 2 white balls in a row.

A32 refers to taking 2 white balls out of 3 white balls, and A52 is taking 2 random balls out of 5 balls.

In fact, it doesn't have to be so complicated, there are only 5 balls 3 white and 2 black, and the first time you take a white ball, there are 4 balls left 2 white rock vertical 2 black, and the general flutter rate of the second time you take a white ball is 2 4 is equal to 1 2

(1) The probability of each ball being taken = [c(1,1)c(9,1)+c(9,1)c(1,1)] a(10,2)=1 5.

2) If a ball cannot be taken in three tests, the probability = (1-1 5) 3 = 64 125, so the probability of being taken = 1-(1-1 5) 3 = 1-64 125 = 81 125

At least one = 100% - all white balls.

c(2 4) denotes the probability of taking two out of four.

That is, 1-1 c(2 4)=1-1 6=5 6 The first red ball c(1 4)*2=1 2, and the second red ball c(1 4)*2=1 2, then the two probabilities are multiplied to get 1 2*1 2=1 4

That is, the probability of two red balls is 1 4

There must be 5 and then there can be 2, 4, 6, 8, one or both of them, 3, 5, 7, 9, up to one of them, with a probability of 11 252

If there's a mistake, it's like I'm making soy sauce.