Probability-Sheldon Ross(Self Doubt)

Question

I was reading probability from Sheldon Ross and found below example.

A purchaser of electrical components buys them in lots of size 10. It is his policy to inspect 3 components from a lot and to accept the lot if all the 3 are nondefective. If 30 percent of the lots have 4 defective components and 70 percent have only 1, what proportion of lots does the purchaser reject?

And the solution is presented like this

Let A denote the event that the purchaser accepts a lot. Now,

P(A)= $P(A|lot\,has\,4\,defectives\,)\frac{3}{10}+P(A|lot\,has\,1\,defective)\frac{7}{10}$

I understood the conditional probability part but why they have multiplied it with $\frac{3}{10}$ in the first part and $\frac{7}{10}$ in the second?

Comments

why because in 30% he have 4 defectives and 70% he have 1 defective

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why not @Ayush

is answer 19/50?

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$\frac{3}{10}\times \frac{4}{5}+\frac{7}{10}\times \frac{1}{5}$

$=\frac{19}{50}$

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@srestha, how you calculated the values $\frac{4}{5} and \frac{1}{5} $

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total number of faults 5

among which 30% have 4 fault

right?

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@Shaik  If we expand the conditional probability then we get P(A∩Lot having 4 defs)/P(lot having 4 defs).

Here the denominator itself is 3/10 isn't it?

So multiplying it with 3/10 will cancel the effect of 3/10. Correct me where I am going wrong..

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@MiniPanda, 

we are dividing as two parts one is 30% which having 4 defective components and another is 70% which having 1 defective components

note that i am calculating A where

A denote the event that the purchaser accepts a lot. Now,

P(A) =  ( P( A | lot has 4 defectives ) * (3/10)  ) +  ( P(A | lot has 1 defective) * ( 7/10 ) )

     but   P( A | lot has 4 defectives ) =  0 due to which ever you taken it has 4 defectives

∴ P(A) = ( P(A | lot has 1 defective) * ( 7/10 ) )

required = 1 - P ( A ) 

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That means there is no probability that the lot containing 4 defectives be accepted? But why so? There are 10 components in each lot and the rest 6 components in that lot are good and they should also contribute to the acceptance of the lot.

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@MiniPanda, yes you are correct.... i confused with no.of lots and size of the lots

 

P( A | lot has 4 defectives ) = $\frac{\binom{6}{3}}{\binom{10}{3}}$

P( A | lot has 1 defectives ) = $\frac{\binom{9}{3}}{\binom{10}{3}}$

 

P(A) =  ( $\frac{1}{6}$  * $\frac{3}{10} $) +  ( $\frac{7}{10}$  * $\frac{7}{10} $ )

      =  ( $\frac{1}{20}$ ) +  ( $\frac{49}{100}$)

      =  ( $\frac{5}{100}$ ) +  ( $\frac{49}{100}$)

      =  ( $\frac{54}{100}$ )

 

required = 1- ( $\frac{54}{100}$ ) =  ( $\frac{46}{100}$ )

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"( 1/6  * 3/10) +  ( 7/10  * 7/10 )"

how??

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Oh yes!!

@Ayush Upadhyaya

We know P(A)=P(A/E₁)*P(E₁) + P(A/E₂)*P(E₂) where E_i for all i are mutually exclusive events in the sample space S and A is any other event in S intersecting with each E_i.

here, P(E₁)=P(lot has 4 defectives)= probability of picking a lot having 4 defectives= 30/100=3/10,

P(E₂)=P(lot has 1 defective)=70/100=7/10

P(A/E₁)= Probability of accepting a lot having 4 defectives

P(A/E₂)= Probability of accepting a lot having 1 defective

P(A)= P(A|lot has 4 defectives)*P(lot has 4 defectives)+P(A|lot has 1 defective)*P(lot has 1 defective)

The rest is like Shaik's.

 

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@ srestha mam,

 

P( A | lot has 4 defectives ) = C( 6, 3 ) / C( 10, 3 ) due to 

for selecting a lot you should have choose three good components.....

in that group only 6 of a lot are good ===> you can choose any three from that six

but total possibility =  you can choose any three from that ten components

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final answer is 46%

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Yes.. Shaik did a small mistake in calculating 7*7. Then the answer is coming 46.

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I updated my comment...

Oh god, save me from calculation mistakes..

Thanks @MiNiPanda