A) \[1/13\]
B) \[3/13\]
C) \[2/13\]
D) None of the above
Correct Answer: C
Solution :
Case I. When zero ace \[{{P}_{1}}({{x}_{1}}=0)=\frac{48}{52}\times \frac{48}{52}={{\left( \frac{12}{13} \right)}^{2}}\] Case II. When one ace \[{{P}_{2}}({{x}_{2}}=1)=\frac{4}{52}\times \frac{48}{52}+\frac{48}{52}\times \frac{4}{52}\] \[=\frac{12}{{{(13)}^{2}}}+\frac{12}{{{(13)}^{2}}}=\frac{24}{{{(13)}^{2}}}\] Case III. When two aces \[{{P}_{3}}({{x}_{3}}=2)=\frac{4}{52}\times \frac{4}{52}={{\left( \frac{1}{13} \right)}^{2}}\] Now, mean \[={{P}_{1}}{{x}_{1}}+{{P}_{2}}{{x}_{2}}+{{P}_{3}}{{x}_{3}}\] \[={{\left( \frac{12}{13} \right)}^{2}}\times 0+\frac{24}{{{(13)}^{2}}}\times 1+{{\left( \frac{1}{13} \right)}^{2}}\times 2\] \[=\frac{24+2}{169}=\frac{26}{169}=\frac{2}{13}\]
You need to login to perform this action.
You will be redirected in
3 sec
