First to 10!

The following cases result in flipping the two black cards before red sums to 10 or more:

Case 1: 2 cards: B5, B5 = $\frac{2}{7}\cdot\frac{1}{6}=\frac{1}{21}$

Case 2: 3 cards: B5, any Red, B5 = $\frac{2}{7}\cdot\frac{5}{6}\cdot\frac{1}{5}=\frac{1}{21}\cdot2=\frac{2}{21}$ (multiply by the number of permutations of the first two elements)

Case 3: 4 cards: B5, any Red, any Red, B5 = $\frac{2}{7}\cdot\frac{5}{6}\cdot\frac{4}{5}\cdot\frac{1}{4}=\frac{1}{21}\cdot3=\frac{1}{7}$

Case 4: 5 cards: 6 sub-cases:

B5, R1, R2, R3, B5
B5, R1, R2, R4, B5
B5, R1, R2, R5, B5
B5, R1, R3, R4, B5
B5, R1, R3, R5, B5
B5, R2, R3, R4, B5

$\frac{2}{7}\cdot\frac{1}{6}\cdot\frac{1}{5}\cdot\frac{1}{4}\cdot\frac{1}{3}\cdot4!\cdot6=\frac{4}{35}$ (4! due to the permutations of the first 4 cards, 6 due to the 6 cases, all of which have the same probability)

Add all these probabilities and simplify = $\frac{2}{5}$

$2+5=7$

There are $\binom{7}{2} = 21$ places the black cards could be in the deck. We consider the possible places for the second black 5 in the deck. It could be in position 2, 3, 4, 5, 6, or 7.

If it is in position 2, 3, or 4, then the sum of the black cards will be 10 first. This has a probability of $\frac{1+2+3}{21} = \frac{2}{7}$ .

If it is in position 6 or 7, then the sum of the red cards will be 10 first. This happens with probability $\frac{5 + 6}{21} = \frac{11}{21}$ .

This leaves a $1 - \frac{2}{7} - \frac{11}{21} = \frac{4}{21}$ chance that the second black card is in the 5th position. If the sum of the red cards after this is at least 6, then the black cards will reach 10 first, otherwise the red cards will. There are $\binom{5}{2} = 10$ ways to choose the cards that come after the second 5. Of these, $(1,5)$ , $(2,4)$ , $(2,5)$ , $(3,4)$ , $(3,5)$ , $(4,5)$ have a total of at least 6. So the probability will be $\frac{6}{10} = \frac{3}{5}$ that the black reaches 10 first. So the total probability of this occurring is $\frac{3}{5}\times \frac{4}{21} = \frac{4}{35}$ .

Thus, the probability that black reaches 10 first is $\frac{2}{7} + \frac{4}{35} = \frac{14}{35} = \frac{2}{5}$ . So $a + b = 7$ .

required P(E)= 2/5 , 2+5 = 7