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 B be the event that the second component drawn is defective.

(a) Find P(A)
(b) Find P(B | A)
(c) Find P(B)

(a)

(b)

(c)

Step-by-step explanation:

Given , Number of components in a lot = 10

Number of  defective components = 3

Let be the event that the first component drawn is defective, and let be the event that the second component drawn is defective.

Probability that  first component drawn is defective=

After first component is drawn , the total component left =9

defective component left = 2

Probability that the second component is defective given that first one is also defective :

Probability that second component drawn is defective = (First component good , second is defective+ First is defective , second is also defective)

hello :

70y^8 + 30y^6 = 2 ( 35 y^8 + 15y^6)

the lenght is : 35 y^8

the width is : 15y^6