Answer to Question #239426 in Statistics and Probability for Bbb

Question #239426
Before the distribution of gadgets, the teacher will test two tablets. Use D for defective tablet, Nfor
non-defective tablet.
Let X be the random variable representing the number of defective tablet that will occur.
Complete the table below and find the values of random variable X.
1
Expert's answer
2021-09-20T18:14:39-0400
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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