Question: Li has $t$ toy bricks. She only has red bricks and blue bricks. Li picks two bricks, one after the other. If the first brick she picks is red, the probability that the second brick is red is $\frac{2}{3}$. If the first brick she picks is blue, the probability that the second brick is red is $\frac{7}{10}$. Calculate the value of $t$.

1. **State the problem:** Li has $t$ toy bricks consisting of red and blue bricks. We want to find $t$ given the probabilities of picking red bricks in two draws. 2. **Define variables:** Let $R$ be the number of red bricks and $B$ be the number of blue bricks. We know: $$R + B = t$$ 3. **Use the given probabilities:** - Probability second brick is red given first is red: $$P(\text{2nd red} | \text{1st red}) = \frac{R-1}{t-1} = \frac{2}{3}$$ - Probability second brick is red given first is blue: $$P(\text{2nd red} | \text{1st blue}) = \frac{R}{t-1} = \frac{7}{10}$$ 4. **Write equations from probabilities:** From the first: $$\frac{R-1}{t-1} = \frac{2}{3}$$ Multiply both sides by $t-1$: $$R - 1 = \frac{2}{3}(t - 1)$$ From the second: $$\frac{R}{t-1} = \frac{7}{10}$$ Multiply both sides by $t-1$: $$R = \frac{7}{10}(t - 1)$$ 5. **Set the two expressions for $R$ equal:** From above: $$R - 1 = \frac{2}{3}(t - 1)$$ $$R = \frac{7}{10}(t - 1)$$ Substitute $R$ from second into first: $$\frac{7}{10}(t - 1) - 1 = \frac{2}{3}(t - 1)$$ 6. **Solve for $t$:** Multiply both sides by 30 (LCM of 10 and 3) to clear denominators: $$30 \times \left(\frac{7}{10}(t - 1) - 1\right) = 30 \times \frac{2}{3}(t - 1)$$ Simplify: $$3 \times 7 (t - 1) - 30 = 10 \times 2 (t - 1)$$ $$21(t - 1) - 30 = 20(t - 1)$$ Expand: $$21t - 21 - 30 = 20t - 20$$ $$21t - 51 = 20t - 20$$ Bring terms to one side: $$21t - 20t = -20 + 51$$ $$t = 31$$ 7. **Final answer:** $$\boxed{31}$$ So, the total number of bricks $t$ is 31.