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.


Expert's answer

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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