Probability tree diagram showing the process of transferring balls between two bags and calculating the likelihood of specific color outcomes.

(a) Complete the tree diagram shown below.

- First branch (from start): Red (1/2), Yellow (1/2)
- From Red:
- Second branch: Red (2/5) → Third branch: Red (5/9), Yellow (4/9)
- Second branch: Yellow (3/5) → Third branch: Red (5/9), Yellow (4/9)
- From Yellow:
- Second branch: Red (3/5) → Third branch: Red (5/9), Yellow (4/9)
- Second branch: Yellow (2/5) → Third branch: Red (4/9), Yellow (5/9)

(b) Find P(A)

P(A) = P(both balls from P are red) + P(both balls from P are yellow)
= (1/2)(2/5) + (1/2)(2/5)
= 1/5 + 1/5
= 2/5

(c) Show that P(B) = 5/9

P(B) = P(third ball is red)
= P(Red, Red, Red) + P(Red, Yellow, Red) + P(Yellow, Red, Red) + P(Yellow, Yellow, Red)
= (1/2)(2/5)(5/9) + (1/2)(3/5)(5/9) + (1/2)(3/5)(5/9) + (1/2)(2/5)(4/9)
= (1/2)(5/9)[(2/5) + (3/5) + (3/5)] + (1/2)(2/5)(4/9)
= (1/2)(5/9)(8/5) + (1/2)(8/45)
= (1/2)(8/9) + (4/45)
= 4/9 + 4/45
= 20/45 + 4/45
= 24/45
= 8/15

Wait — correction needed. Let’s recalculate P(B) carefully.

Actually, after first two draws from P, bag Q has 9 balls. The number of red balls in Q depends on what was added:

Case 1: Two reds added to Q → Q has 6 red, 3 yellow → P(red) = 6/9 = 2/3
Case 2: One red, one yellow added → Q has 5 red, 4 yellow → P(red) = 5/9
Case 3: Two yellows added → Q has 4 red, 5 yellow → P(red) = 4/9

Now compute:

P(B) = P(two reds from P) * P(red | two reds added) + P(one red, one yellow) * P(red | mixed added) + P(two yellows from P) * P(red | two yellows added)

P(two reds from P) = (3/6)(2/5) = 1/5
P(two yellows from P) = (3/6)(2/5) = 1/5
P(one red, one yellow) = 1 - 1/5 - 1/5 = 3/5

So,

P(B) = (1/5)(6/9) + (3/5)(5/9) + (1/5)(4/9)
= (1/5)(2/3) + (3/5)(5/9) + (1/5)(4/9)
= (2/15) + (15/45) + (4/45)
= (2/15) + (1/3) + (4/45)
Convert to 45 denominator:
= 6/45 + 15/45 + 4/45
= 25/45
= 5/9 ✓

(d) Show that P(A ∩ B) = 2/9

A: both balls from P same color → either both red or both yellow.
B: third ball is red.

So A ∩ B = (both red from P and third red) OR (both yellow from P and third red)

P(both red from P and third red) = P(first red) * P(second red | first red) * P(third red | two reds added)
= (3/6)(2/5)(6/9) = (1/2)(2/5)(2/3) = (1/5)(2/3) = 2/15

P(both yellow from P and third red) = P(first yellow) * P(second yellow | first yellow) * P(third red | two yellows added)
= (3/6)(2/5)(4/9) = (1/2)(2/5)(4/9) = (1/5)(4/9) = 4/45

So P(A ∩ B) = 2/15 + 4/45 = 6/45 + 4/45 = 10/45 = 2/9 ✓

(e) Hence find P(A ∪ B)

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
= 2/5 + 5/9 - 2/9
= 2/5 + 3/9
= 2/5 + 1/3
= 6/15 + 5/15
= 11/15

(f) Given that all three balls drawn are the same colour, find the probability that they are all red.

Let C = event that all three balls are same color.

C = (all red) OR (all yellow)

P(all red) = P(first red) * P(second red | first red) * P(third red | two reds added)
= (3/6)(2/5)(6/9) = (1/2)(2/5)(2/3) = 2/15

P(all yellow) = P(first yellow) * P(second yellow | first yellow) * P(third yellow | two yellows added)
= (3/6)(2/5)(5/9) = (1/2)(2/5)(5/9) = (1/5)(5/9) = 1/9

P(C) = P(all red) + P(all yellow) = 2/15 + 1/9 = 6/45 + 5/45 = 11/45

We want P(all red | C) = P(all red) / P(C) = (2/15) / (11/45) = (2/15)*(45/11) = (2*3)/11 = 6/11