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The mean of defective blades supplied in packets of $10$ is $1$. in how many packets of this make out of $1000$ packages would you expect to find at least $4$ non defective blades.

Answer given in my book is $13$ whereas me and my friends are getting $999$

Henry
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Tanay Joshi
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  • Have you tried using the binomial distribution mentioned in the title to your question? – Henry Mar 21 '19 at 13:10
  • Yes. Answer given in my book is 13 whereas me and my friends are getting 999 – Tanay Joshi Mar 21 '19 at 13:11
  • I would double check that you have the problem written down correctly. If the question asks for the number that have at least $4$ non defectives, then your answer would be right. But if it's asking for at least $4$ defectives, then the book's answer is right. – WaveX Mar 21 '19 at 13:51
  • Its asking for 4 non defectives. So is 999 correct in this case – Tanay Joshi Mar 21 '19 at 13:56
  • $13$ or more precisely $12.7951984$ is the expected number of packages with at least $4$ defective blades - in R this would be (1-pbinom(3,10,0.1))*1000 – Henry Mar 21 '19 at 20:11
  • Question is 4 NON DEFECTIVE – Tanay Joshi Mar 22 '19 at 02:18

2 Answers2

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To get you started:

The mean number of defective blades in a pack of $10$ is $1$, meaning the probability that a given blade in a pack is defective is $0.1$.

What is the probability that in $10$ blades, at least $4$ of them are defective?

Hint: Think Binomial Distribution and Complement Rule.

Then use this answer, which will be the probability that a given pack will have at least $4$ defective blades, and multiply by $1000$ to find the expected number of packs that have at least $4$ defective blades.

WaveX
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p = 0.1 q = 0.9 n = 10 N = 1000 (x is no. of defective items)

P(X=x) + 10Cx * (0.1)^x * (0.9)^(10-x)

P(X>=4) = 1 - P(X=0,1,2,3 non defective items)

0 non defective = 10 defective 1 non defective = 9 defective and so on

1 - P(X=0,1,2,3 non defective items) = 1 - (10C10 * (0.1)^10 * (0.9)^0 + 10C9 * (0.1)^9 * (0.9)^1 + 10C8 * (0.1)^8 * (0.9)^2 + 10C7 * (0.1)^7 * (0.9)^3) = 0.999

N*0.999 = 999

Tanay Joshi
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