A lumber company has just taken delivery on a shipment of 10,000 2 ✕ 4 boards. Suppose that 10% of these boards (1000) are actually too green to be used in first-quality construction. Two boards are selected at random, one after the other. Let A = {the first board is green} and B = {the second board is green}. (a) Compute P(A), P(B), and P(A ∩ B) (a tree diagram might help). (Round your answer for P(A ∩ B) to five decimal places.)

(b) With A and B independent and P(A) = P(B) = 0.1, what is P(A ∩ B)?

Step-by-step explanation:

Given That:

Let A = {the first board is green} and B = {the second board is green}.

A lumber company has just taken delivery on a shipment of 10,000 2×4 boards.

Suppose that 10% of these boards (1000) are actually too green to be used in first-quality construction. Two boards are selected at random, one after the other.

We are to compute the following probabilities :

P(A)

P(B)

P(A ∩ B)

To start with the probability P(A)

P(A) = 0.1

where;

Recall that :

P(A ∩ B) = P(B|A)B

P(A ∩ B) = 0.0999 × 0.10

P(A ∩ B) = 0.00999

(b)

Given that A and B are independent; Then:

P(A ∩ B) = P(A) × P(B)

0.00999 = 0.1 × 0.09999

0.00999 =  0.00999

As such A and B are independent

However; when P(A ∩ B) = P(A) = P(B) = 0.1

P(A ∩ B) = P(A) × P(B)

P(A ∩ B) = 0.1 × 0.1

P(A ∩ B) =  0.01