Answer:
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
