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Biscuits are sold in packets of 18. There is a constant probability that any biscuit is...

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jspake | Student, Undergraduate | eNoter

Posted April 21, 2013 at 8:17 AM via web

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Biscuits are sold in packets of 18. There is a constant probability that any biscuit is broken, independently of other biscuits. The mean number of broken biscuits in a packet has been found to be 2.7. Find the probability that a packet contains between 2 and 4 (inclusive) broken biscuits.

Please help me solve the question above.. I'm totally confused. An explanation will be appreciated!
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Tagged with math, probability, statistics

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jspake | Student , Undergraduate | eNoter

Posted April 21, 2013 at 12:54 PM (Answer #2)

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Thanks

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pramodpandey | College Teacher | Valedictorian

Posted April 21, 2013 at 10:07 AM (Answer #1)

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Let random selection of broken biscuit follows Binomial distribtion.

mean of binomial distribution= np

2.7=np

But n=18

2.7=18p

p=2.7/18

p=3/20

=.15

q=1-p

=.85

Probability of r broken biscuits.

P(x=r)=`c(n,r)p^rq^(n-r)`

`Thus`

`` `P(x=2)=C(18,2)(.15)^2(.85)^16`

`P(x=3)=C(18,3)(.15)^3(.85)^15`

`P(x=4)=C(18,4)(.15)^4(.85)14`

probability that a packet contains between 2 and 4 (inclusive) broken biscuits  = P(x=2)+P(x=3)+P(x=4)

ans.

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