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

move it to the other side

step-by-step explanation: