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From a lot of 6 items containing 2 defective items, a sample of 4 items are drawn at random. Let the random variable $X$ denote the number of defective items in the sample. If the sample is drawn without replacement, find Mean of X. \[\]

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Last updated date: 20th Jun 2024
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Answer
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Hint: We see that the random variable $X$ can take the values $X=0,X=1,X=2$since there can be 0 or 1 or 2 defective items from the selected sample. We find $P\left( X=0 \right)$ by finding a number of ways to select 0 defective items from 2 defective items simultaneously 4 defective terms from 6 defective items. We similarly find $P\left( X=1 \right),P\left( X=2 \right)$. We find the mean as $\sum{{{x}_{i}}P\left( X={{x}_{i}} \right)}$ where$i=1,2,3$.\[\]

Complete step-by-step solution:
We are given that From a lot of 6 items containing 2 defective items, a sample of 4 items are drawn at random. We can select out defective items in ${}^{6}{{C}_{4}}$ ways which is the total number of outcomes. \[\]
We are further given that the random variable $X$ denotes the number of defective items in the sample and the sample is drawn without replacement. So if there is no defective item in the sample$X=0$, if there is one defective item in the sample $X=1$ and if there is two defective items in the sample $X=2$ .\[\]
We select 0 defective items from 2 defective item in ${}^{2}{{C}_{0}}$ way and 4 non-defective items from 4 non-defective items in ${}^{4}{{C}_{4}}$ ways. So by rule of product the number of ways we can select 4 items where 0 items are defective is ${}^{2}{{C}_{0}}\times {}^{4}{{C}_{1}}$. So the probability that there is 0 defective items in the sample is
\[P\left( X=0 \right)=\dfrac{{}^{2}{{C}_{0}}\times {}^{4}{{C}_{4}}}{{}^{6}{{C}_{4}}}=\dfrac{1\times 1}{15}=\dfrac{1}{15}\]
We select 1 defective item from 2 defective item in ${}^{2}{{C}_{1}}$ way and 3 non-defective items from 4 non-defective items in ${}^{4}{{C}_{3}}$ ways. So by rule of product the number of ways we can select 4 items where 1 item is defective is ${}^{2}{{C}_{1}}\times {}^{4}{{C}_{3}}$. So the probability that there is 1 defective item in the sample is
\[P\left( X=1 \right)=\dfrac{{}^{2}{{C}_{1}}\times {}^{4}{{C}_{3}}}{{}^{6}{{C}_{4}}}=\dfrac{2\times 4}{15}=\dfrac{8}{15}\]
 We select 2 defective items from 2 defective items in ${}^{2}{{C}_{2}}$ way and 2 non-defective items from 2 non-defective items in ${}^{4}{{C}_{4}}$ ways. So by rule of product the number of ways we can select 4 items where 2 items are defective is ${}^{2}{{C}_{2}}\times {}^{4}{{C}_{2}}$. So the probability that there is 2 defective items in the sample is
\[P\left( X=2 \right)=\dfrac{{}^{2}{{C}_{2}}\times {}^{4}{{C}_{2}}}{{}^{6}{{C}_{4}}}=\dfrac{1\times 6}{15}=\dfrac{6}{15}\]
 So the expectation mean of the random variable $X$ is
\[\begin{align}
  & E\left( X \right)=0\cdot P\left( X=0 \right)+1\cdot P\left( X=1 \right)+2\cdot \left( X=2 \right) \\
 & \Rightarrow E\left( X \right)=0\cdot \dfrac{1}{15}+1\cdot \dfrac{8}{15}+2\cdot \dfrac{6}{15} \\
 & \Rightarrow E\left( X \right)=\dfrac{8}{15}+\dfrac{12}{15}=\dfrac{20}{15}=\dfrac{4}{3} \\
\end{align}\]

Note: We note that we can select $r$ objects from $n$ distinct objects without replacement in ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ ways otherwise with replacement in ${}^{n+r-1}{{C}_{r}}$ ways. We also note that the mean or expected value of a random variable that has finite outcomes ${{x}_{1}},{{x}_{2}},...,{{x}_{k}}$ with probabilities ${{p}_{1}},{{p}_{2}},...,{{p}_{k}}$ is given by $E\left( X \right)=\sum{{{x}_{i}}{{p}_{i}}}$and the variance is given by $Var\left( X \right)=E\left( {{X}^{2}} \right)-{{\left( E\left( X \right) \right)}^{2}}$.