Question #175864

Ten android tablets will be delivered to a certain school for testing but three of them are defective. The principal will get two of these tablets for testing. Let X represent the defective tablet, Y represent the non-defective tablet and L represent the random variable representing the defective tablet will occur. Complete the table below and find the values of random variable L.


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Expert's answer
2021-03-29T12:59:49-0400
Possible Values of the OutcomesRandom Variable LLL2LY1YL1YY0\def\arraystretch{1.5} \begin{array}{c:c} \text{Possible } & \text{Values of the } \\ \text{Outcomes} & \text{Random Variable} \ L\\ \hline LL & 2 \\ LY & 1 \\ YL & 1\\ YY & 0 \end{array}

P(L=0)=(30)(10320)(102)=2145=715P(L=0)=\dfrac{\dbinom{3}{0}\dbinom{10-3}{2-0}}{\dbinom{10}{2}}=\dfrac{21}{45}=\dfrac{7}{15}

P(L=1)=(31)(10321)(102)=2145=715P(L=1)=\dfrac{\dbinom{3}{1}\dbinom{10-3}{2-1}}{\dbinom{10}{2}}=\dfrac{21}{45}=\dfrac{7}{15}

P(L=2)=(32)(10322)(102)=345=115P(L=2)=\dfrac{\dbinom{3}{2}\dbinom{10-3}{2-2}}{\dbinom{10}{2}}=\dfrac{3}{45}=\dfrac{1}{15}

l012P(L=l)715715115\def\arraystretch{1.5} \begin{array}{c:c} l & 0 & 1 & 2 \\ \hdashline \\ P(L=l) & \dfrac{7}{15} & \dfrac{7}{15} & \dfrac{1}{15} \end{array}




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