Suppose that you have 12 identical balls and 3 empty boxes. You randomly put every ball into some box. What is the probability that none of the boxes have more than 6 balls?

the probability that none of the boxes have more than 6 balls is 0.3077

Step-by-step explanation:

Given that;

12 balls are put into 3 boxes randomly, without ay condition

so we will be using the multinomial formula;

⇒ [ 12 + 3 - 1              [ 14

        3 - 1 ]         =       2 ]    =    91

now, assuming that one of the box has more than 6 balls that is at least 7 balls

x + y + z = 12

 x + y + z = 7

therefore

x + 7 + y + z = 12

x + y + z = 5

therefore the number of the solution here computed as;

⇒ [ 5 + 3 - 1              [ 7

        3 - 1 ]         =       2 ]    =   21

Hence, the probability that none of the boxes have more than six (6) balls will be;

= (91 - (3 × 21)) / 91

= (91 - 63) / 91

= 28 / 91

= 0.3077

Therefore the probability that none of the boxes have more than 6 balls is 0.3077

could you reply to this with the question so i can ?

22, 28, 34, 40, 46

b)   f(n) = 6n + 16

n = what is the number of the row ( 22 would be row 1) so if we replaced n with 1, we would get 22

c)   the 14th row will have 100 seats

(1)   (2)   (3)   (4)   (5)   (6)   (7)   (8)   (9)   (10) (11) (12) (13) (14)

22 | 28 | 34 | 40 | 46 | 52 | 58 | 64 | 70 | 76 | 82 | 88 | 94 | 100