A) \[\frac{35}{47}\]
B) \[\frac{55}{1024}\]
C) \[\frac{220}{512}\]
D) \[\frac{11}{204}\]
Correct Answer: B
Solution :
We have, \[p(X=r){{=}^{5}}{{C}_{r}}{{\left( \frac{1}{2} \right)}^{5}}\] And \[P(Y=r){{=}^{7}}{{C}_{r}}{{\left( \frac{1}{2} \right)}^{7}}\] Now, \[P(X+Y=3)=P(X=0,Y=3)+P(X=1,Y=2)\] \[=P(X=2,Y=1)+P(X=3,Y=0)\] \[=P(X=0).P(Y=3)+P(X=1).P(Y=2)\] \[+P(X=2).P(Y=1)+P(X=3).P(Y=0)\]\[{{=}^{5}}{{C}_{0}}{{\left( \frac{1}{2} \right)}^{5}}{{\times }^{7}}{{C}_{3}}{{\left( \frac{1}{2} \right)}^{7}}{{+}^{5}}{{C}_{1}}{{\left( \frac{1}{5} \right)}^{5}}{{\times }^{7}}{{C}_{2}}{{\left( \frac{1}{2} \right)}^{7}}\] \[{{+}^{5}}{{C}_{2}}{{\left( \frac{1}{2} \right)}^{5}}{{\times }^{7}}{{C}_{1}}{{\left( \frac{1}{2} \right)}^{7}}{{+}^{5}}{{C}_{3}}{{\left( \frac{1}{5} \right)}^{5}}{{\times }^{7}}{{C}_{0}}{{\left( \frac{1}{2} \right)}^{7}}\] \[=\frac{1}{{{2}^{12}}}[35+5\times 21+10\times 7+10\times 1]\] \[=\frac{220}{1024\times 4}=\frac{55}{1024}\]
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