Answer:(a) [tex]P(A)=0.3[/tex]
(b)[tex]P(B|A)=\dfrac{2}{9}[/tex]
(c) [tex]P(B)=0.3[/tex]
Step-by-step explanation:
Given , Number of components in a lot = 10
Number of defective components = 3
Let [tex]A[/tex] be the event that the first component drawn is defective, and let [tex]B[/tex] be the event that the second component drawn is defective.
Probability that first component drawn is defective=[tex]P(A)=\dfrac{3}{10}=0.3[/tex]
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 : [tex]P(B|A)=\dfrac{\text{Number of bad components left}}{\text{Total components left}}=\dfrac{2}{9}[/tex]
Probability that second component drawn is defective = (First component good , second is defective+ First is defective , second is also defective)
[tex]P(B)=\dfrac{7}{10}\times\dfrac{3}{9}+\dfrac{3}{10}\times\dfrac{2}{9}=\dfrac{3}{10}=0.3[/tex]
