Marbles of Probability

Let $x$ be the probability of drawing a red ball.

P(3R,2B)= (no. of ways to arrange 3R and 2B in order)( $x^3$ )( $1-x)^2$

$\therefore , \frac {5!}{3!\cdot 2!}\cdot x^3(1-x)^2=32\cdot \frac {5!}{1!4!}\cdot x(1-x)^4$

Solving this equation will give you $x=0.8$ .

Therefore, percentage of red balls in the bag = 80%

Let p be the probability of drawing a red marble.

In that case, $1-p$ is the probability of drawing a blue marble.

The probability of drawing 3 red marbles and 2 blue marbles is ${5 \choose 3} \times {p^3} \times {(1-p)^2}$ . ${5 \choose 3}=10$ .

The probability of drawing 1 red marble and 4 blue marbles is $5 \times p \times {(1-p)^4}$ .

Since the probability of drawing 3 red marbles and 2 blue marbles is 32 times the probability of drawing 1 red marble and 4 blue marbles, we can form the equation $10{p^3}{(1-p)^2}={32 \times 5p{(1-p)^4}}$ .

We need to solve for p. To do that, we can first divide both sides by $10p{(1-p)^2}$ , and get ${p^2}=16{(1-p)^2}$ .

We then distribute the right side of the equation to get ${p^2}=16-32p+16{p^2}$ .

Next, we subtract ${p^2}$ from both sides, to further simplify our equation to $0=15{p^2}-32p+16$ .

Finally, we can use the quadratic formula to solve for p: $p=\frac{-(-32) \pm \sqrt{(-32)^2-4(15)(16)}}{2(15)}$ .

If we add $\sqrt{(-32)^2-4(15)(16)}$ , p will be greater than one, so we can only subtract it, to achieve $p=0.8$ . However, the answer needs to be written as a percentage, so we need to multiply p by 100, to get our final answer of $\boxed{80}$ .

Edit: Brian Charlesworth mentioned a short-cut for solving for p. View the comments on this solution to see. Note: p cannot equal $-4(1-p)$ as that would make p greater than one.

$\text{Interpreter}[\text{PercentFraction}][\text{Select}[p\text{/.}\, \text{Solve}[\text{PDF}[\text{BinomialDistribution}[5,p]][3]=32\, \text{PDF}[\text{BinomialDistribution}[5,p]][1]],0<\#<1\&][[1]]] \Rightarrow 80$ %.

The heart of the solution is the Solve function with the Binomial distribution parameterized by p and equating the probability of 3 red marbles to 32 times the probability of 1 red marble. The Select function removes the nonsense solutions for $p$ as $p$ has to be strictly between $0$ and $1$ and not equal to either $0$ or $1$ as both red and blue marbles may be drawn. The Interpreter function returns a function that converts the value from the Select function call, $\frac45$ to a textual 80%.

$\text{PDF}\left[\text{BinomialDistribution}\left[5,\frac{4}{5}\right]\right] \Rightarrow \text{Function}\left[x, \begin{array}{cc} \{ & \begin{array}{cc} \frac{4^x \binom{5}{x}}{3125} & 0\leq x\leq 5 \\ 0 & \text{True} \\ \end{array} \\ \end{array} \right]$

In other words, the PDF function returns an anonymous function of one argument, the number of red marbles.

The anonymous functions returns $\frac{128}{625}$ for 3 red marbles and $\frac{4}{625}$ for 1 red marble when p is $\frac45$ .

The equation that Solve solves is $10 (1 - p)^2 p^3 == 160 (1 - p)^4 p$ which simplifies to $15 p^4-47 p^3+48 p^2-16 p=0$ .

That solution is $\left\{p\to 0,p\to \frac{4}{5},p\to 1,p\to \frac{4}{3}\right\}$ . Zero and one can be eliminated for reasons given above. The fraction $\frac43$ can be eliminated as $0\leq p\leq 1$ for statistical reasons.The remaining solution is the answer.