Question

....

Answer 1

I have:
Part a) 1/6
Part b) 8/9

Answer 2

a) (probability of selecting bushy trees from A) * (probability of selecting lean trees from B)

Answer 3

b) It's a little bit more complicated as you need to consider multiple cases.

- 2 bushy trees from A, 2 bushy trees from B, 1 bushy from A + 1bushy from B

Answer 4

Did i crush the numbers correct, @SAValkyrie .
I am not quite sure about part b.

Answer 5

Tell how how you got 1/6 for A

Answer 6

P(B|X) = 1/2,
P(L|y) = 1/3,
1/2 * 1/3 = 1/6

Answer 7

Oh no, i missed a part, it's two from row a and two from row b.

Answer 8

Picking 2 bushy trees from part A is not 1/2. Your logic is slightly flawed. Think about it

4 bushy trees out of 8 trees.

Picking 1 bushy tree from A: 4/8
Picking 2nd bushy tree from A : 3/7

Probability of picking 2 bushy trees : 4/8 * 3/7 =3/14

Answer 9

Yep. 3/14 + (3/9 * 2/8) = 25/84.

Answer 10

You need to multiply probability from A with probability from B since they need to occur at the same time to satisfy the question.

It should be 3/14 * ( 3/9 * 2/8) ?

Answer 11

Oh right.

Answer 12

For exactly two to be bushy,
(4/8 * 3/7) * (6/9 * 5/8), correct?

Answer 13

@Directrix , what are your thoughts here?