anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
okay so i know i have to find P(x greaterthanorequalto 1)
anonymous
  • anonymous
@jim_thompson5910 @Astrophysics @pooja195 @ganeshie8
jim_thompson5910
  • jim_thompson5910
at least one? is that what you meant?

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anonymous
  • anonymous
yes
jim_thompson5910
  • jim_thompson5910
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
anonymous
  • anonymous
yes
jim_thompson5910
  • jim_thompson5910
so we can say P(A) = 1 - P(B) we just need to figure out P(B)
anonymous
  • anonymous
so the probability that no calculators are defective?
jim_thompson5910
  • jim_thompson5910
yes
jim_thompson5910
  • jim_thompson5910
how many non-defective calcs are there?
anonymous
  • anonymous
37 out of 55 which is 67% chance of one not being defective
jim_thompson5910
  • jim_thompson5910
there are 37 non defective calcs how many ways are there to pick 4 of them (order doesn't matter) ?
anonymous
  • anonymous
24?
jim_thompson5910
  • jim_thompson5910
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
jim_thompson5910
  • jim_thompson5910
since order doesn't matter, you divide that by 4! = 24
anonymous
  • anonymous
okay that's interesting what i did was multiply 4*3*2
anonymous
  • anonymous
anyway what do i do now?
jim_thompson5910
  • jim_thompson5910
what do you get when you divide by 24?
anonymous
  • anonymous
do you mean divide the 67% ?
jim_thompson5910
  • jim_thompson5910
divide 1,585,080 by 24
anonymous
  • anonymous
okay 66045 , what does the 66045 represent?
jim_thompson5910
  • jim_thompson5910
it represents the number of ways to pick 4 calculators out of the pool of 37
anonymous
  • anonymous
okay okay so we now multiply this by the probability to get a non defective calc right? and then subtract that from 1?
jim_thompson5910
  • jim_thompson5910
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
jim_thompson5910
  • jim_thompson5910
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) = ???
anonymous
  • anonymous
.19936
anonymous
  • anonymous
and we subtract that from 1 right?
jim_thompson5910
  • jim_thompson5910
I'm getting (66045)/(341055) = 0.19364911817742
anonymous
  • anonymous
woops one too many nines , yes im getting that too
anonymous
  • anonymous
and yep the answer is .806
anonymous
  • anonymous
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?
anonymous
  • anonymous
how can i think more like you do?
jim_thompson5910
  • jim_thompson5910
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
jim_thompson5910
  • jim_thompson5910
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 \[\Large C(n,r) = \frac{n!}{r!(n-r)!}\]
anonymous
  • anonymous
alright i need practice , thanks for you help really appreciate it :D
jim_thompson5910
  • jim_thompson5910
no problem

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