# Question #2c069

##### 1 Answer

(please excuse the long answer)

#### Explanation:

We know that

This kind of probability problem is all about finding an equivalent problem which is easier to solve but has the same solution.

**PART 1:**

Let's think about it this way: we have 9 balls that we need to distribute into 4 boxes labeled "a", "b", "c", and "d". The number of balls in each box determines the value of the variable on its label. How many different ways can we distribute these balls, given that the balls are indistinguishable?

This seems difficult to calculate, but we can make it easier by thinking about the boxes as just a way to separate the integers from one another. This way, we can represent the number of balls in all four boxes as a string of letters. It may seem confusing at first, but bear with me...

Let's call each ball "b"

Let's call each separation between boxes "s"For example, if there are 3 balls in A, 2 balls in B, no balls in C, and 4 balls in D, then the string of letters will look like this:

#color(blue)(b" " b " "b)" " s" " color(blue)(b " "b)" " s" " s" "color(blue)( b" " b" " b" " b#

Now, all we have to do is find the total number of ways that we can arrange 9 balls and 3 "separations", given that both of these are indistinguishable.

Since we could get the same result by placing 3

#s# 's in three random spots, and then filling the remaining 9 spots with#b# 's, we can find the total number of ways to fill all 12 spots by finding the total number of combinations of picking three positions out of 12.This is just "12 choose 3", or:

#""_12C_3 = (12!)/(9!*3!) = (12 xx 11 xx 10)/6 = 2 xx 11 xx 10 = 220#

Therefore, there are 220 ways to separate 9

Therefore, there are 220 ways to place 9 identical balls in 4 labeled boxes.

Therefore, there are 220 different sums of 4 non-negative numbers which equal 9.

**PART 2:**

We need to find the probability that

If

We can use the same "balls in boxes" analogy as in part 1 here: we have 7 balls to put in 3 boxes, which means we have 7

This means that our total number of combinations is "9 choose 2":

#""_9C_2 = (9!)/(7!*2!) = (9 xx 8)/2 = 9 xx 4 = 36#

Therefore, there are 36 different sums of 4 numbers which equal 9, given that the first number (

**PART 3:**

The probability that

#P(a=2) = 36/220 = 9/55#

*Final Answer*