Planet Brilliantia

This is a conditional probability problem. Let $A$ be the event that the first creature's statement was the truth and $B$ be the event that the second creature says that the first creature's statement was true. We know that

$P(A|B) = \frac{P(A \cap B)}{P(B)}.$

We will find $P(B)$ and $P(A \cap B)$ . In order to do that, we must split into four cases, based on the type of creature the two are. Then, we split each case into two sub-cases based on whether the first creature was telling the truth or not. (If the first creature told the truth, the second creature also told the truth. If the first creature lied, the second also lied.) For brevity, I will condense all these cases into a single table.

$\begin{array}{c|c|c|c} \text{Type of first} & \text{Type of second} & \text{First's statement} & \text{Probability} \\ \hline \text{Math} & \text{Math} & \text{True} & \frac{2}{3} \cdot \frac{6}{7} \cdot \frac{2}{3} \cdot \frac{6}{7} \\ \text{Math} & \text{Non-math} & \text{True} & \frac{2}{3} \cdot \frac{6}{7} \cdot \frac{1}{3} \cdot \frac{1}{5} \\ \text{Non-math} & \text{Math} & \text{True} & \frac{1}{3} \cdot \frac{1}{5} \cdot \frac{2}{3} \cdot \frac{6}{7} \\ \text{Non-math} & \text{Non-math} & \text{True} & \frac{1}{3} \cdot \frac{1}{5} \cdot \frac{1}{3} \cdot \frac{1}{5} \\ \hline \text{Math} & \text{Math} & \text{False} & \frac{2}{3} \cdot \frac{1}{7} \cdot \frac{2}{3} \cdot \frac{1}{7} \\ \text{Math} & \text{Non-math} & \text{False} & \frac{2}{3} \cdot \frac{1}{7} \cdot \frac{1}{3} \cdot \frac{1}{5} \\ \text{Non-math} & \text{Math} & \text{False} & \frac{1}{3} \cdot \frac{4}{5} \cdot \frac{2}{3} \cdot \frac{1}{7} \\ \text{Non-math} & \text{Non-math} & \text{False} & \frac{1}{3} \cdot \frac{4}{5} \cdot \frac{1}{3} \cdot \frac{4}{5} \\ \end{array}$

The probability $P(B)$ represents the sum of all the probabilities in the rightmost column, while the probability $P(A \cap B)$ represents the sum of all the "true" probabilities i.e. the first four. The sum of the first four probabilities equals $\left ( \frac{67}{105} \right ) ^2$ , and the sum of all the probabilities equals $\left ( \frac{67}{105} \right ) ^2 + \left ( \frac{38}{105} \right ) ^2$ . Thus,

$\begin{aligned} P(A|B) &= \frac{P(A \cap B)}{P(B)} \\ &= \frac{\left ( \frac{67}{105} \right ) ^2}{ \left ( \frac{67}{105} \right ) ^2 + \left ( \frac{38}{105} \right ) ^2} \\ &= \frac{67^2}{67^2 + 38^2}, \end{aligned}$

and $b - a = (67^2 + 38^2) - 67^2 = 38^2 = \boxed{1444}$ .

The probability the any statement said on Brilliantia is true is $\frac {2}{3} \cdot \frac {6}{7} + \frac {1}{3} \cdot \frac {1}{5} = \frac {67}{105}$ .

The probability the any statement said on Brilliantia is a false is $1 - \frac {67}{105} = \frac {38}{105}$ .

If the first statement is true, and the second creature says it's true, the second creature must be telling the truth. The probability of that is $(\frac {67}{105})^2 = \frac {4489}{11025}$ .

If the first statement is false, and the second creature says it's true, the second creature must be lying. The probability of that is $(\frac {38}{105})^2 = \frac {1444}{11025}$ .

So the probability of the first statement is true, given the second creature says it's true is $\frac {\frac {4489}{11025}}{\frac {4489}{11025} + \frac {1444}{11025}} = \frac {4489}{5933}$ .

$5933 - 4489 = \boxed {1444}$

probability of 1st creature saying truth is (2/3)(6/7)+(1/3)(1/5), you can use this and its counterpart directly and reach last step of calculation.