Question

find the indicated probability

a sample of 4 different calculators is randomly selected from a group containing 18 that are defective and 37 that have no defects. What is the probability that at least one of the calculators is defective?

Answer 1

okay so i know i have to find P(x greaterthanorequalto 1)

Answer 2

@jim_thompson5910 @Astrophysics @pooja195 @ganeshie8

Answer 3

at least one? is that what you meant?

Answer 4

yes

Answer 5

do you agree that

P(A) + P(B) = 1

where

A = the event that at least one calculator is defective
B = the event that no calculators are defective

Answer 6

yes

Answer 7

so we can say

P(A) = 1 - P(B)

we just need to figure out P(B)

Answer 8

so the probability that no calculators are defective?

Answer 9

yes

Answer 10

how many non-defective calcs are there?

Answer 11

37 out of 55 which is 67% chance of one not being defective

Answer 12

there are 37 non defective calcs

how many ways are there to pick 4 of them (order doesn't matter) ?

Answer 13

24?

Answer 14

slot 1: 37 choices
slot 2: 36 choices
slot 3: 35 choices
slot 4: 34 choices

IF order mattered, then you'd have 37*36*35*34 = 1,585,080 different ways to select 4 of the 37 non defective calcs

Answer 15

since order doesn't matter, you divide that by 4! = 24

Answer 16

okay that's interesting what i did was multiply 4*3*2

Answer 17

anyway what do i do now?

Answer 18

what do you get when you divide by 24?

Answer 19

do you mean divide the 67% ?

Answer 20

divide 1,585,080 by 24

Answer 21

okay 66045 , what does the 66045 represent?

Answer 22

it represents the number of ways to pick 4 calculators out of the pool of 37

Answer 23

okay okay so we now multiply this by the probability to get a non defective calc right?
and then subtract that from 1?

Answer 24

we do the same thing but now we include the 18 defective calcs

so we have 18+37 = 55 total

there are C(55,4) = 341,055 ways to pick 4 calcs (defective or nondefective) out of the pool of 55


The notation C(n,r) is the nCr combination notation

Answer 25

B = the event that no calculators are defective


P(B) = (# of ways to pick 4 non defective calculators)/(# of ways to pick any 4 calculators)

P(B) = (66,045)/(341,055)

P(B) = ???

Answer 26

.19936

Answer 27

and we subtract that from 1 right?

Answer 28

I'm getting

(66045)/(341055) = 0.19364911817742

Answer 29

woops one too many nines , yes im getting that too

Answer 30

and yep the answer is .806

Answer 31

damn okay so how do you know what to do , i get that its logic and all but is their some special thought process that you go through to find the answer?

Answer 32

how can i think more like you do?

Answer 33

I think the first thing to determine is that you have to understand that

P(A) + P(B) = 1

where

A = the event that at least one calculator is defective
B = the event that no calculators are defective

events A and B are complementary events. Only one event must occur

Answer 34

which leads to P(A) = 1 - P(B)

to find P(B), you need to compute both

C(37, 4)
C(55, 4)

You use the combination formula each time

\[\Large C(n,r) = \frac{n!}{r!(n-r)!}\]

Answer 35

alright i need practice , thanks for you help really appreciate it :D

Answer 36

no problem