in how many ways can you throw<randomly>, 3 balls into 10 boxes if each of the boxes can hold all 3 balls?

Question

my problem here is that i wish to find the probabilities of a box containing one,two or three balls.
I request a detailed solution and a possible illustration of the probability distribution if a random variable X is the number of boxes with atleast one ball.

Answer 1

Each ball can be found in any of the 10 boxes.
So the number of combinations is

10 for the first ball x
10 for the second ball x
10 for the third ball
= 1000

The probability P1 that a box holds ONE ball is the number of combinations where it has one ball divided by 1000.
So put one ball into the box, and count the combinations of 2 balls in the 9 other boxes: 9*9 = 81 Hence P1 = 81/1000

Similarly P2 is 9/1000

and P3 is 1/1000

Answer 2

Since each box can hold 3 balls, there are 10 possible boxes for each ball that you throw; therefore, there are 10^3=1000 different ways in which the balls can be distributed. The probability of getting three balls in a single box would be 1/10*1/10=1/100; the probability of getting two balls in a single box would be 2*9/10*1/10=18/100; the chance of having all three balls land in different boxes is 9/10*9/10=81/100.

Answer 3

If the bins can carry 6 balls and each and each be distinctive, then considered one of them ought to be 1x6x1, yet another 2x3x1 flat, and a third, 3x2x1 vertical (on end) yet as long via fact the bins are saved orientated in only one way - say one outer area is white dealing with the consumer/digicam and the precise open, then each and each container has a similar form of combinations. (which does no longer be actual in the event that they could be grew to become approximately so purple-Blue grew to become into comparable to Blue-purple) 6*5*4*3*2*a million or 720 in keeping with container and 2160 for the three bins.

Answer 4

10*10*10= 1000 diffrent ways