Michael's marbles

The probability of drawing a red marble first and a blue marble second is $\frac{R}{R+B} \cdot \frac{B}{R+B} = \frac{RB}{(R+B)^2}.$ This is also the probability of drawing a blue marble first and a red marble second. So the probability of drawing one marble of each color is $\frac{2RB}{(R+B)^2} = \frac{15}{32}.$ Now note that $\begin{aligned} \frac{R}{B} + \frac{B}{R} &= \frac{R^2+B^2}{RB} \\ &= \frac{(R+B)^2 - 2RB}{RB} \\ &= 2 \frac{(R+B)^2}{2RB} - 2 \\ &= 2 \cdot \frac{32}{15} - 2 = \frac{34}{15}. \end{aligned}$ Therefore, $a + b = 34 + 15 = \boxed{49}.$

Note that the values of $R$ and $B$ are not uniquely determined by the conditions of the problem. In particular, explicitly solving the quadratic relation in $R, B$ yields $R/B = 3/5$ or $R/B = 5/3$ .

We are asked to find the value of $\frac{R}{B} + \frac{B}{R}$ . By expressing as one fraction.

$\frac{R^{2} + B^{2}}{RB}$

We can express the probability of the each marble of different color be drawn as, $\frac{RB}{(R + B)^{2}}$ and if you picked the blue marble first, $\frac{BR}{(R + B)^{2}}$ . Since the events above are independent of each other, adding them will lead to the probability stated in the problem.

From the given:

$\frac{2RB}{(R + B)^{2}} = \frac{15}{32}$

by cross multiplication:

$64 RB = 15(R+B)^{2}$

$64 RB = 15R^{2} + 30RB + 15B^{2}$

$34 RB = 15(R^{2} + B^{2})$

Which can be expressed as in the ratios as above.

$\frac{34}{15} = \frac{R^{2} + B^{2}}{RB}$

Therefore, $34 + 15 = 49$

First, it's helpful to rewrite $\frac{R}{B} + \frac{B}{R}$ as $\frac{R^2 + B^2}{RB}$ .

Next, we set up an equation for the probability of drawing one marble of each color:

$2*\frac{R}{R+B}*\frac{B}{R+B} = \frac{2RB}{(R+B)^2} = \frac{15}{32}$

Cross-multiply to get $64RB = 15R^2 + 30RB + 15B^2$ , or $34RB = 15(R^2 + B^2)$ .

Divide both sides by $15RB$ , and you get $\frac{R^2 + B^2}{RB} = \frac{34}{15}$ , so $a+b=49$ .

The total number of marbles in the box is clearly $B+R$ . Because we are working with replacement, the probability that one marble of each color is drawn is $\binom{2}{1}\frac{B}{B+R}\frac{R}{B+R}$ , since there are $\binom{2}{1}$ ways of picking which marble you draw out first, the probability you pick a blue marble is $\frac{B}{B+R}$ , and the probability you pick a red marble is $\frac{R}{B+R}$ . This gives us that $\frac{2BR}{(B+R)^2} = \frac{15}{32}$ .

We cross-multiply to get that $64BR = 15B^2 + 30BR + 15R^2$ . Combining like terms and dividing through by $15BR$ gives us that $\frac{34}{15} = \frac{B^2+R^2}{BR} = \frac{B}{R} + \frac{R}{B}$ . Thus, our desired quantity is $15 + 34 = \fbox{49}$ .

We first calculate the probability $P$ of drawing a red marble and a blue marble. We find

$P=(\frac{R}{R+B})(\frac{B}{R+B})+(\frac{B}{R+B})(\frac{R}{R+B})=\frac{15}{32}\rightarrow \frac{R\cdot B}{R^2+B^2+2R\cdot B}=\frac{15}{64}$

We are looking for $\frac{R}{B}+\frac{B}{R}$ ; this can be simplified to $\frac{R^2+B^2}{R\cdot B}$ . This is similar to our expression for $P$ , so we try to simplify that expression:

$\frac{R\cdot B}{R^2+B^2+2R\cdot B}=\frac{15}{64}\rightarrow\frac{64}{15}=\frac{R^2+B^2+2R\cdot B}{R\cdot B}$

So now we have

$\frac{64}{15}=\frac{a}{b}+\frac{2R\cdot B}{R\cdot B}$

where $a$ and $b$ are the constants we are told to find. This implies

$\frac{34}{15}=\frac{a}{b}$ .

$34$ and $15$ are coprime, so our answer is $34+15=\boxed{49}$

total number of marbles = R+B

so chance of drawing one of each with replacement is R/(R+B) * B/(R+B)+B/(R+B) * R/(R+B)=15/32

we have RB/(R+B)^2=15/64

we can deduce R=3 and B=5

R/B+B/R=3/5+5/3=34/15

hence the answer is 34+15=49

Probability of getting two Reds = (R/(R+B))*(R/(R+B))

Probability of getting two Blues = (B/(R+B))*(B/(R+B))

Probability of getting one Red and one Blue

1- [( R R + B B)/ (R+B)*(R+B)]

= [(R+B)(R+B) - (R R + B B)]/[(R+B)(R+B)]

= 2 R B/[(R+B)*(R+B)] = 15/32 = 2. 15/ [(5+3)(5+3)]

So R/B + B/R = (5/3) + (3//5)

             = 34/15 = a/b

a+b= 34+25 =49

Clearly, Required probability = $2(\frac{R}{R + B})(\frac{B}{R + B}) = \frac{15}{32} \\ \Rightarrow \frac{(R + B)^2}{RB} = \frac{64}{15} \\ \Rightarrow \frac{R^2 +B^2 + 2RB}{RB} = \frac{64}{15} \\ \Rightarrow \frac{R}{B} +\frac{B}{R} + 2 = \frac{64}{15} \\ \Rightarrow \frac{R}{B} +\frac{B}{R} = \frac{34}{15} = \frac{a}{b} \\ \Rightarrow a + b = 49$

We know that if we want to select two different colored marbles, We can have first, a red marble $R$ and then we can have a blue marble $B$ or we can have a blue marble $B$ and then a red marble $R$ . So we get the probability that the two marbles drawn are different as $2\cdot\frac{R}{R+B}\cdot\frac{B}{R+B}=\frac{2RB}{R^2+2RB+B^2}=\frac{15}{32}.$ Now we see that we have $2RB=15,$ thus $R^2+B^2=32-15=17.$ (Note that these aren't necessarily the values of $R$ and $B$ but these will still give us the right answer since the ratios are still all the same.)

We notice that we can express $\frac{R}{B}+\frac{B}{R}$ as $\frac{R^2+B^2}{RB}.$ Right here we recall that we have $R^2+B^2=17$ and we also have $2RB=15$ thus $RB=\frac{15}{2}.$ So $\frac{R}{B}+\frac{B}{R}=\frac{34}{15}$

Now we finally have found our answer of $a+b=\boxed {49}$

Let r be the probability of choosing a red marble and b be the probability of choosing a blue marble. the probability of choosing both can be split into two cases, where both of them are $rb$ . Therefore, $2rb=\frac{15}{32} \implies rb=\frac{15}{64}$ .

We also know that $r+b=1$ , so a little guess and check leaves us with $r=\frac{5}{8} \text{and} b=\frac{3}{8}$ . Note that we can switch around the values because of the question. From here, we can say there are 5 red marbles and 3 blue marbles, or 3 red marbles and 5 blue marbles.

Either way the answer is still $\frac{5}{3}+\frac{3}{5}=\frac{34}{15} \implies \boxed{49}$

2[RB/(R+B)^2]=15/32

[RB/(R+B)^2]=15/64

[RB/(R+B)^2]=(3*5)/(3+5)^2

Thus, R=3 and B=5, or R=5 and B=3.

R/B+B/R=3/5+5/3=34/15=a/b

Therefore, a+b=49

The probability that one marble of each color is drawn is equal to 2(R)(B)/(R+B)^2 = 15/32, or (R)(B)/(R+B)^2 = 15/64, R/B + B/R = (R^2+B^2)/(B R) = (R+B)^2 / (B R) - 2 = 64/15 - 2 = 34/15. So a+b = 49

2BR/(B+R)^2=15/32. So (B+R)^2/BR=(B^2+R^2)/RB + 2 = 64/15.

Means that R/B + B/R = 64/15 - 2 = 34/15.

So a+b=34+15=49.