A lumber company has just taken delivery on a shipment of 10,000 2 âœ• 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 âˆ© B) (a tree diagram might help). (Round your answer for P(A âˆ© B) to five decimal places.)

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

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)

[tex]P(A) = \dfrac{1000}{10000}[/tex]

P(A) = 0.1

[tex]P(B) = P(B|A)*P(A)+P(B|A')*P(A')[/tex]

where;

[tex]P(B|A) = \dfrac{N(B|A)}{N-1}[/tex]

[tex]P(B|A) = \dfrac{999}{9999}[/tex]

[tex]P(B|A) =0.0999[/tex]

[tex]P(B|A') = \dfrac{N(B|A')}{N-1}[/tex]

[tex]P(B|A') = \dfrac{1000}{999}[/tex]

[tex]P(B|A') = 0.10[/tex]

Recall that :

[tex]P(B) = P(B|A)*P(A)+P(B|A')*P(A')[/tex]

[tex]P(B) = 0.0999*0.1+0.10*(1-0.1)[/tex]

[tex]P(B) = 0.00999+0.10*(0.9)[/tex]

[tex]P(B) = 0.00999+0.09[/tex]

[tex]P(B) = 0.0999[/tex]

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