Question #288495

A shipment of 20 similar laptop computers to a retail outlet contains 3 that are defective. If a school makes a random purchase of 2 of these computers, find the probability distribution for the number of defectives. Show the table in your solution. Round off answers to 4 decimal places.


1
Expert's answer
2022-01-19T03:22:38-0500

Let X=X= the number of defective computers purchased: X=0,1,2.X=0,1,2.


P(X=0)=(30)(20320)(202)P(X=0)=\dfrac{\dbinom{3}{0}\dbinom{20-3}{2-0}}{\dbinom{20}{2}}

=1(136)190=6895=\dfrac{1(136)}{190}=\dfrac{68}{95}


P(X=1)=(31)(20321)(202)P(X=1)=\dfrac{\dbinom{3}{1}\dbinom{20-3}{2-1}}{\dbinom{20}{2}}

=3(17)190=51190=\dfrac{3(17)}{190}=\dfrac{51}{190}



P(X=2)=(32)(20322)(202)P(X=2)=\dfrac{\dbinom{3}{2}\dbinom{20-3}{2-2}}{\dbinom{20}{2}}

=3(1)190=3190=\dfrac{3(1)}{190}=\dfrac{3}{190}

The probability distribution for the number of defectives


x012p(x)0.71580.26840.0158\def\arraystretch{1.5} \begin{array}{c:c} x & 0 & 1 & 2 \\ \hline p(x) & 0.7158 & 0.2684 & 0.0158 \\ \end{array}



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