Probability tree diagram illustrating the outcomes and probabilities of selecting two balls from a set, where B = blue and R = red.

Let’s solve this step by step.

We are given a probability tree diagram for picking two items (with replacement? or without? — we’ll check from the numbers).

From the diagram:

- First pick:
- P(B) = 7/10 → blue
- P(R) = 3/10 → red

Then, second pick depends on first pick? Let’s look at the branches.

After first pick is B (blue), second pick has:
- P(B | first was B) = 6/9 → so outcome BB
- P(R | first was B) = ? Not written, but since only two outcomes, it must be 3/9? Wait — actually, in the diagram, after first B, second pick shows 6/9 for B, and then R branch doesn’t have a number — but logically, if total probability must add to 1, then P(R | first B) = 1 - 6/9 = 3/9.

Similarly, after first pick R, second pick:
- Branches to B and R, but no probabilities shown. But again, they should add to 1. However, looking at the structure, maybe the second pick probabilities depend on what was picked first — suggesting sampling WITHOUT replacement.

Wait — let’s think: If you pick a blue first (out of 10 total?), then for second pick, there are 9 left, and 6 blues remain? That would mean originally there were 7 blues and 3 reds (since 7+3=10). Then if you pick one blue first, you have 6 blues and 3 reds left → total 9. So yes, that matches: P(second B | first B) = 6/9.

Similarly, if first pick is red (3 reds out of 10), then after removing one red, you have 7 blues and 2 reds left → total 9. So P(second B | first R) = 7/9, and P(second R | first R) = 2/9.

But in the diagram, those numbers aren't filled in for the R-first branch. However, to compute all outcome probabilities, we need them.

So let’s reconstruct the full tree with correct conditional probabilities.

Total items: 10 (7 blue, 3 red)

First pick:
- P(B) = 7/10
- P(R) = 3/10

Second pick (without replacement):

If first was B:
- Remaining: 6B, 3R → total 9
→ P(B|B) = 6/9
→ P(R|B) = 3/9

If first was R:
- Remaining: 7B, 2R → total 9
→ P(B|R) = 7/9
→ P(R|R) = 2/9

Now, compute each outcome probability by multiplying along the branches.

Outcome BB:
P(BB) = P(first B) × P(second B | first B) = (7/10) × (6/9) = 42/90 = 7/15 ≈ 0.4667

Outcome BR:
P(BR) = P(first B) × P(second R | first B) = (7/10) × (3/9) = 21/90 = 7/30 ≈ 0.2333

Outcome RB:
P(RB) = P(first R) × P(second B | first R) = (3/10) × (7/9) = 21/90 = 7/30 ≈ 0.2333

Outcome RR:
P(RR) = P(first R) × P(second R | first R) = (3/10) × (2/9) = 6/90 = 1/15 ≈ 0.0667

Check: Total probability = 42/90 + 21/90 + 21/90 + 6/90 = 90/90 = 1 → Good.

The question didn’t explicitly ask for anything specific — but since the user said “solve the problem”, and the image shows a tree with missing probabilities, likely the task is to fill in the missing probabilities for each outcome.

Looking back at the image: under “Probability” column, all are blank. So probably the task is to calculate the probability of each outcome: BB, BR, RB, RR.

So final answers:

P(BB) = 42/90 = simplify: divide numerator and denominator by 6 → 7/15
But maybe leave as 42/90 or reduce fully? Usually reduce.

42 ÷ 6 = 7, 90 ÷ 6 = 15 → 7/15
21/90 = 7/30
21/90 = 7/30
6/90 = 1/15

Alternatively, keep denominator 90 for consistency? But simplified is better.

Since the original fractions used denominators 10 and 9, perhaps express over 90.

But let’s see what’s expected. In school problems, often simplified fractions are preferred.

So:

BB: 7/15
BR: 7/30
RB: 7/30
RR: 1/15

Double-check calculations:

(7/10)*(6/9) = 42/90 = 7/15 ✔️
(7/10)*(3/9) = 21/90 = 7/30 ✔️
(3/10)*(7/9) = 21/90 = 7/30 ✔️
(3/10)*(2/9) = 6/90 = 1/15 ✔️

All good.

Final Answer:
P(BB) = \frac{7}{15}, P(BR) = \frac{7}{30}, P(RB) = \frac{7}{30}, P(RR) = \frac{1}{15}