Balls And Bags

There are 90! possible orders in which the individual balls can be drawn. Of these, there are $20 \times 89!$ possible sequences in which the 15th ball is red, $30 \times 89!$ sequences in which it's white and $40 \times 89!$ in which it's blue. This probability distribution is the same whether we are looking at the 1st ball drawn, the 15th ball drawn, the 90th ball drawn or any other.

The total number of sequences is not changed by changing the order in which balls are drawn.

You don't need to do any complex combinatorics to solve this problem, because there's a very simple symmetry argument here. Label the balls 1 through 90. For every ball i=1,2,3...90, the probability that you draw ball i on the 15th draw is 1/90.

There are 20 balls that are red. Whatever numbers i these have, each of these red balls -- like every ball -- has probability 1/90 of being drawn on the 15th draw. Thus the probability that the 15th draw is a red ball is 20/90.

Incredibly, the probability of drawing any color ball after $n$ balls are drawn is the same as the probability of drawing that color ball WITH replacement! The probability of drawing a red ball on the $n$ th draw with replacement is $\frac{20}{90}=\frac{2}{9}$ , therefore our answer must be $2+9=\boxed{11}$ .

If you didn't find Garrett Clarke 's slick and effortless solution, you still could've solved it. Here is another solution that does not involve with replacement.

Notice that white and blue are arbitrary so we can include them as just one color. Let's call this color grey. Therefore there are 20 red balls and 70 grey balls.

We do casework on the number of red balls.

1 red ball: ${14 \choose 0}70 \cdot 69 \cdot ... \cdot 57 \cdot 20 = {14 \choose 0} \frac{70!}{56!} \cdot \frac{20!}{19!}$

2 red balls: ${14 \choose 1}70 \cdot 69 \cdot 68 \cdot ... \cdot 58 \cdot 20 \cdot 19 = {14 \choose 1} \frac{70!}{57!} \cdot \frac{20!}{18!}$

$\vdots$

15 red balls: ${14 \choose 14} 20 \cdot 19 ... \cdot 6 = {14 \choose 14}\frac{70!}{70!} \cdot \frac{20!}{5!}$

Thus the total number of ways is $\begin{aligned} A & = {14 \choose 0}\frac{70!}{56!} \cdot \frac{20!}{19!} + {14 \choose 1}\frac{70!}{57!} \cdot \frac{20!}{18!} + ... + {14 \choose 14}\cdot\frac{70!}{70!} \cdot \frac{20!}{5!} \\ & = \frac{70! \cdot 20!}{75!}\left({14 \choose 0}{75 \choose 19} + {14 \choose 1}{75 \choose 18} + ... + {14 \choose 14}{75 \choose 5}\right)\\ & = \frac{70! \cdot 20!}{75!}B \end{aligned}$

Now we are stuck, how do we simplify $B$ ? Before we expand the binomial coefficients, let's analyze the sum. Notice that if we have $14$ boys and $75$ girls , $B$ is just the number of ways to form a subcommittee of 19 people. However, this is equivalent to $B = {14 + 75 \choose 19} = {89 \choose 19}$ .

Therefore, $\begin{aligned} A & = \frac{70! \cdot 20!}{74!}{89 \choose 19} = \frac{70! \cdot 20! \cdot 89!}{19! \cdot 70! \cdot 75!} = \frac{20 \cdot 89!}{75!} \end{aligned}$

We know that the total number of ways to pick 15 is $C = 90 \cdot 89 \cdot ... \cdot 76 = \frac{90!}{75!}$ .

Therefore the probability is just $\frac{A}{C}$ . $\begin{aligned} \frac{A}{C} & = \frac{\frac{20 \cdot 89!}{75!}}{\frac{90!}{75!}} = \frac{20}{90} = \frac{2}{9} \implies p+q = \boxed{11} \end{aligned}$

However this solution is long and the most desirable solution is noticing that counting with replacement keeps the probability the same.

Whatever ball that we draw, it is random and we do not necessarily know the color of the ball that we have drawn, we are only dealing with the 15th ball. Since by drawing the balls, we do not gain any new information on the color of the ball, the probability of drawing a red ball will remain the same throughout which is $\frac{20}{90}$ = $\frac{2}{9}$

I got the right answer by calculating the expected number of red balls after 14 were taken out. The expected number of reds taken out was therefore 14 * $\frac{2}{9}$ = $\frac{28}{9}$ . This means the expected number of red balls left is 20- $\frac{28}{9}$ = $\frac{152}{9}$ . so the probability of getting a red ball for the last draw is $\frac{152}{9}$ divided by (90-14) which is $\frac{2}{9}$

Divide the process in two steps: First, grab 15 balls at random. Then put these balls in random order.

Originally the frequency of red balls is $2/9$ , The sample of 15 balls has the same expected value for the frequency of red balls.

Given a random order, the probability that the last ball is red is equal to the expected value for the frequency of red balls.

Thus the answer is $2/9$ , or $\boxed{11}$ .