Question 159205
Please read the comments:

4.5 ( 2, 16, 32) 
2. How many ways can a basebal manager arrange a batting order of 9 player? 
I did 9!= 362,880 please let me know if this is correct.

GOOD WORK
 
16. An inspector must select 3 tests to perform in a certain order on a manufactured part. He has a choice of 7 tests. How many ways can he perform 3 different tests?
I did 7P3 = 7!/(7-3)! = 7!/4! = 210 Please let me know if this is correct.

GOOD WORK
 
32. If a person can select 3 presents from 10 presents under a christmas tree, how many different combinations are there?
I did 10C3 = 10!/(10-3)!3!= 10!/7!3!= 10*9*8*7!/7!*3*2*1= 10*3*4=120 
Please let me know if this is correct. 

GOOD WORK
Section 4-6 problem number 6 
6. A package contains 12 resistors, 3 of which are defective. If 4 are selected, fin the probability of getting

a. No defective resistors
b. 1 Defective resistor
c. 3 Defective resistors 
for a. P(no defectives)= 9C4/12C4=126/495 

GOOD WORK

and then I got confused. 
for b P(1 defective) = 1- P(no defectives) (WRONG)
= 1- 9C4/12C4
= 1- 126/595
Then I get confuse becaue I'm not sure how to
subtract from the one. 

Answer:
P(1 defective)=P(1 def, 3 no def)=(3C1)(9C3)/12C4 =252/495 =28/55


for c. P(3 defective) = 3C3*9C3/12C4 = 94/495 I think i'm supposed to reduce it but im not sure how or if this is correc. (WRONG)

Solution:

P(3 defective)=P(3 def, 1 no def)=(3C3)(9C1)/12C4=9/495=1/55