An urn holds 5 red balls and 3 white balls…

Step 1. Both red. The first ball is red with probability $\dfrac{5}{8}$. Then the urn holds $7$ red and $3$ white balls, so the second is red with probability $\dfrac{7}{10}$. Both red: $\dfrac{5}{8} \times \dfrac{7}{10} = \dfrac{7}{16} = 0.4375$.

Step 2. The second ball is red. The other path is white then red. The first ball is white with probability $\dfrac{3}{8}$. Then the urn holds $5$ red and $5$ white balls, so red has probability $\dfrac{5}{10}$. That path gives $\dfrac{3}{8} \times \dfrac{5}{10} = \dfrac{3}{16}$. Adding the two paths: $\dfrac{7}{16} + \dfrac{3}{16} = \dfrac{10}{16} = 0.625$.

Step 3. First red, given second red. $P = \dfrac{7/16}{10/16} = \dfrac{7}{10} = 0.7$.