Calculating Binomial Probability for Car Color Preferences

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SUMMARY

The discussion focuses on calculating binomial probabilities related to car color preferences, specifically determining the likelihood of selecting between three and five black cars from a sample of 20, where the probability of a car being black is 10%. The correct approach involves calculating P(3 ≤ x ≤ 5), which includes P(x = 3), P(x = 4), and P(x = 5). The final probability is confirmed to be 0.312, correcting the initial miscalculation that only considered P(3 < x ≤ 5).

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tzx9633

Homework Statement



Car colour preferences changes over year . In this year , suppose that 10% of the car are randomly selected , let the sample of cars are 20 . Find the probabilities between thre and five cars ( inclusive ) are black ...

I am aksed to do this question using binomial .

Homework Equations

The Attempt at a Solution


I treated it as P ( 3 < x ≤ 5 ) , am i right ?
so , P( x = 4) + P ( x = 5)
=
(20c4) ((0.1)^4) ( 0.9^16) + (20c5) ((0.1)^5) ( 0.9^15) = 0.1217 ,

However , the ans given is = 0.312 , which part of my working is wrong ?
 
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tzx9633 said:
I treated it as P ( 3 < x ≤ 5 ) , am i right ?
so , P( x = 4) + P ( x = 5)
=
(20c4) ((0.1)^4) ( 0.9^16) + (20c5) ((0.1)^5) ( 0.9^15) = 0.1217 ,

It is supposed to be P( 3 ≤ x ≤ 5 ). For some reason you treated the '5' correct but not the '3'.

So it will be P( x = 3) + P (x = 4) + P ( x = 5 ) = 0.312
 
solved !
 

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