A) \[6/7\]
B) \[36/49\]
C) \[5/7\]
D) \[25/49\]
Correct Answer: A
Solution :
Given, number of red balls \[=3\] and white balls \[=4\] Case I. When number of red ball is zero. \[{{P}_{1}}({{x}_{1}}=0)=\frac{4}{7}\times \frac{4}{7}\times \frac{4}{7}=\frac{64}{{{(7)}^{3}}}\] Case II. When number of red balls is one. \[{{P}_{2}}({{x}_{2}}=1)=3\left( \frac{3}{7}\times \frac{4}{7}\times \frac{4}{7} \right)=\frac{144}{{{(7)}^{3}}}\] Case III. When number of red balls is two. \[{{P}_{3}}({{x}_{3}}=2)=3\left( \frac{3}{7}\times \frac{3}{7}\times \frac{4}{7} \right)=\frac{108}{{{(7)}^{3}}}\] Case IV. When number of red balls is three. \[{{P}_{4}}({{x}_{4}}=3)=\frac{3}{7}\times \frac{3}{7}\times \frac{3}{7}=\frac{27}{{{(7)}^{3}}}\] \[\therefore \] Variance \[=\sum\limits_{i=0}^{3}{{{p}_{i}}x_{i}^{2}}-{{\left( \sum\limits_{i=0}^{3}{{{p}_{i}}x_{i}^{2}} \right)}^{2}}\] \[=\left[ \frac{64}{{{(7)}^{3}}}\times 0+\frac{144}{{{(7)}^{3}}}\times {{(1)}^{2}}+\frac{108}{{{(7)}^{3}}}\times {{(2)}^{2}}+\frac{27}{{{(7)}^{3}}}\times {{(3)}^{2}} \right]\] \[-{{\left[ \frac{64}{{{(7)}^{3}}}\times 0+\frac{144}{{{(7)}^{3}}}\times 1+\frac{108}{{{(7)}^{3}}}\times 2+\frac{27}{{{(7)}^{3}}}\times 3 \right]}^{2}}\] \[=\left[ 0+\frac{144}{343}+\frac{432}{343}+\frac{243}{343} \right]\] \[-{{\left[ 0+\frac{144}{343}+\frac{216}{343}+\frac{81}{343} \right]}^{2}}\] \[=\frac{819}{343}-{{\left( \frac{441}{343} \right)}^{2}}=\frac{280917-194481}{{{(343)}^{2}}}\] \[=\frac{86436}{{{(343)}^{2}}}=\frac{252}{343}=\frac{36}{49}\] Now, standard deviation \[=\sqrt{\operatorname{var}iance}\] \[=\sqrt{\frac{36}{49}}=\frac{6}{7}\]
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