2 identical baskets A and B contain white and red balls. Basket A contains 7 White balls and 3 red balls while basket B contains 5 white balls and 5 red balls. A bag is chosen at random and 2balls picked from it one after another without replacement .
(a) Illustrate this information using a tree diagram
(b). Find the probability that
(i). The 2 balls picked are of the same colour
(ii). The 2 balls picked are of different colours
(iii) Only one of the balls picked is red
(iv). At least one white ball is picked
(a) Tree diagram:
```
A B
/ \ / \
W R W R
/ \ / \ / \ / \
W R W R W R W R
```
(b)
(i) Probability of picking 2 white balls or 2 red balls:
- From basket A: P(2 white balls) = (7/10) * (6/9) = 7/15
- From basket A: P(2 red balls) = (3/10) * (2/9) = 1/15
- From basket B: P(2 white balls) = (5/10) * (4/9) = 5/18
- From basket B: P(2 red balls) = (5/10) * (4/9) = 5/18
Total probability = P(2 white balls) + P(2 red balls) = 7/15 + 1/15 + 5/18 + 5/18 = 23/45
(ii) Probability of picking 1 white ball and 1 red ball:
- From basket A: P(1 white, 1 red) = (7/10) * (3/9) + (3/10) * (7/9) = 42/90
- From basket B: P(1 white, 1 red) = (5/10) * (5/9) + (5/10) * (5/9) = 25/45
Total probability = P(1 white, 1 red) = 42/90 + 25/45 = 29/45
(iii) Probability of picking only 1 red ball:
- From basket A: P(1 red, 1 not red) = (3/10) * (7/9 + 3/9) = 12/30
- From basket B: P(1 red, 1 not red) = (5/10) * (5/9 + 5/9) = 10/18
Total probability = P(1 red, 1 not red) = 12/30 + 10/18 = 11/15
(iv) Probability of picking at least 1 white ball:
- From basket A: P(at least 1 white) = 1 - P(2 red balls) = 1 - 1/15 = 14/15
- From basket B: P(at least 1 white) = 1 - (5/18)^2 = 47/81
Total probability = P(at least 1 white) = P(A) * P(at least 1 white from A) + P(B) * P(at least 1 white from B)
= 0.5 * 14/15 + 0.5 * 47/81 = 107/135.