Question 159205This question is from textbook Elementary Statistics
: 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.
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.
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.
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 and then I got confused.
for b P(1 defective) = 1- P(no defectives)
= 1- 9C4/12C4
= 1- 126/595
Then I get confuse becaue I'm not sure how to
subtract from the one.
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.
This question is from textbook Elementary Statistics
Answer by oscargut(2103)   (Show Source):
You can put this solution on YOUR website! 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
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