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Binomial Distribution of Random Variable

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0

$begingroup$

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 $999$

binomial-distribution

share|cite|improve this question

edited Mar 21 at 20:12

Henry

101k482170

asked Mar 21 at 13:09

Tanay JoshiTanay Joshi

13

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Have you tried using the binomial distribution mentioned in the title to your question?
  $endgroup$
  – Henry
  Mar 21 at 13:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. Answer given in my book is 13 whereas me and my friends are getting 999
  $endgroup$
  – Tanay Joshi
  Mar 21 at 13:11
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  $endgroup$
  – WaveX
  Mar 21 at 13:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Its asking for 4 non defectives. So is 999 correct in this case
  $endgroup$
  – Tanay Joshi
  Mar 21 at 13:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $13$ 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
  $endgroup$
  – Henry
  Mar 21 at 20:11
  
  

  
  
  
  

 | 
show 1 more comment

0

$begingroup$

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 $999$

binomial-distribution

share|cite|improve this question

edited Mar 21 at 20:12

Henry

101k482170

asked Mar 21 at 13:09

Tanay JoshiTanay Joshi

13

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Have you tried using the binomial distribution mentioned in the title to your question?
  $endgroup$
  – Henry
  Mar 21 at 13:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. Answer given in my book is 13 whereas me and my friends are getting 999
  $endgroup$
  – Tanay Joshi
  Mar 21 at 13:11
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  $endgroup$
  – WaveX
  Mar 21 at 13:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Its asking for 4 non defectives. So is 999 correct in this case
  $endgroup$
  – Tanay Joshi
  Mar 21 at 13:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $13$ 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
  $endgroup$
  – Henry
  Mar 21 at 20:11
  
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

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 $999$

binomial-distribution

share|cite|improve this question

edited Mar 21 at 20:12

Henry

101k482170

asked Mar 21 at 13:09

Tanay JoshiTanay Joshi

13

$endgroup$

share|cite|improve this question

edited Mar 21 at 20:12

Henry

101k482170

asked Mar 21 at 13:09

Tanay JoshiTanay Joshi

13

share|cite|improve this question

edited Mar 21 at 20:12

Henry

101k482170

asked Mar 21 at 13:09

Tanay JoshiTanay Joshi

13

edited Mar 21 at 20:12

Henry

101k482170

edited Mar 21 at 20:12

Henry

101k482170

asked Mar 21 at 13:09

Tanay JoshiTanay Joshi

13

asked Mar 21 at 13:09

Tanay JoshiTanay Joshi

13

2 Answers
2

active

oldest

votes

0

$begingroup$

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.

share|cite|improve this answer

answered Mar 21 at 13:21

WaveXWaveX

2,8442822

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I solved in the same way . I need final answer to compare
  $endgroup$
  – Tanay Joshi
  Mar 21 at 13:22
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  May you share your work by editing your question body? Ultimately I am getting the answer of your book (after rounding)
  $endgroup$
  – WaveX
  Mar 21 at 13:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I posted my approach
  $endgroup$
  – Tanay Joshi
  Mar 21 at 13:39
  

  
  
  
  

add a comment | 

0

$begingroup$

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

share|cite|improve this answer

answered Mar 21 at 13:39

Tanay JoshiTanay Joshi

13

$endgroup$

add a comment | 

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0

$begingroup$

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.

share|cite|improve this answer

answered Mar 21 at 13:21

WaveXWaveX

2,8442822

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I solved in the same way . I need final answer to compare
  $endgroup$
  – Tanay Joshi
  Mar 21 at 13:22
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  May you share your work by editing your question body? Ultimately I am getting the answer of your book (after rounding)
  $endgroup$
  – WaveX
  Mar 21 at 13:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I posted my approach
  $endgroup$
  – Tanay Joshi
  Mar 21 at 13:39
  

  
  
  
  

add a comment | 

0

$begingroup$

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.

share|cite|improve this answer

answered Mar 21 at 13:21

WaveXWaveX

2,8442822

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I solved in the same way . I need final answer to compare
  $endgroup$
  – Tanay Joshi
  Mar 21 at 13:22
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  May you share your work by editing your question body? Ultimately I am getting the answer of your book (after rounding)
  $endgroup$
  – WaveX
  Mar 21 at 13:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I posted my approach
  $endgroup$
  – Tanay Joshi
  Mar 21 at 13:39
  

  
  
  
  

add a comment | 

$begingroup$

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.

share|cite|improve this answer

answered Mar 21 at 13:21

WaveXWaveX

2,8442822

$endgroup$

share|cite|improve this answer

answered Mar 21 at 13:21

WaveXWaveX

2,8442822

answered Mar 21 at 13:21

WaveXWaveX

2,8442822

answered Mar 21 at 13:21

WaveXWaveX

2,8442822

0

$begingroup$

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

share|cite|improve this answer

answered Mar 21 at 13:39

Tanay JoshiTanay Joshi

13

$endgroup$

add a comment | 

0

$begingroup$

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

share|cite|improve this answer

answered Mar 21 at 13:39

Tanay JoshiTanay Joshi

13

$endgroup$

add a comment | 

$begingroup$

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

share|cite|improve this answer

answered Mar 21 at 13:39

Tanay JoshiTanay Joshi

13

$endgroup$

share|cite|improve this answer

answered Mar 21 at 13:39

Tanay JoshiTanay Joshi

13

answered Mar 21 at 13:39

Tanay JoshiTanay Joshi

13

answered Mar 21 at 13:39

Tanay JoshiTanay Joshi

13