A lot of 10 components contains 3 that are defective. Two components are drawn at random and tested. Let A be the event that the first component drawn is defective, and let She the event that the second component drawn is defective.

a. Find P(A).

b. Find P(B|A).

c. Find P(A ∩ B).

d. Find P(Ac∩B).

e. Find P(B).

f. Are A and B independent? Explain.

There are 10 components with 3 that are defective:

A = event that the first component is defective

B = event that the second component is defective

b) P(A) = this is the probability that the first component is defective:

= number of defective component/total number of component

P(A) = 3/10 = 0.3

b) P(B|A) = the probability that the second component is defective after the first defective component. Since A has occurred:

= number of defective component left/total number of component left

P(B|A) = 2/9 = 0.22

c) P(A ∩ B) = probability that both are defective i.e. A and B occurring

= 3/10 * 2/9

P(A ∩ B) = 6/90 = 0.06

d) P(Ac ∩ B) = probability that A is not defective and B is defective

P(Ac) = 1 - 3/10 = 7/10 = 0.7

∴ P(Ac ∩ B) =7/10 * 2/9 = 14/90 = 0.16

e) P(B) = probability that the second component is defective

= 2/9 = 0.22

f) B is dependent on A because the defective component is not replaced after the first selection. There would be 9 items left after the first selection of which 2 are defective.

Step-by-step explanation:

the answer is f(x)=2x-1

step-by-step explanation:

step-by-step explanation:

always plug in x. in this case, x equals 4.

i'm going to first solve 4(2+3x)

1) 4(2+3(4)) use pemdas to solve.

i'm going to multiply whats in the parenthesis first, so 3 times 4=12.

2)4 (2+12) i'm then going to multiply 4 by both numbers in the parenthesis (because multiplying comes before adding)

4 times 2= 8.   4 times 12= 48

4) (8+48)= 56 (answer)

i'm then going to solve 8+12x and plug in 4 for x.

1) 8 +12(4)

multiplying comes before adding, so i'm gonna multiply 12 and 4.

12 times 4= 48.

2) 8+48=56

both equal the same!