A) \[y-1=0\]
B) \[x-1=0\]
C) \[x+y-1=0\]
D) \[x-y+1=0\]
Correct Answer: B
Solution :
By total probability theorem, Required probability \[=\frac{1}{2}\times \left( \frac{3}{5}\times 1+\frac{2}{5}\times \frac{1}{2} \right)\] \[+\frac{1}{2}\left( \frac{^{3}{{C}_{2}}}{^{5}{{C}_{2}}}\times 1+\frac{^{2}{{C}_{2}}}{^{5}{{C}_{2}}}\times \frac{1}{3}+\frac{^{3}{{C}_{1}}{{\times }^{2}}{{C}_{1}}}{^{5}{{C}_{2}}}\times \frac{2}{3} \right)\] \[=\frac{1}{2}\left( \frac{3}{5}+\frac{1}{5} \right)+\frac{1}{2}\left( \frac{3}{10}+\frac{1}{30}+\frac{2}{5} \right)\]\[=\frac{2}{5}+\frac{1}{2}\left( \frac{9+1+12}{30} \right)\]\[=\frac{2}{3}+\frac{22}{60}=\frac{24+22}{60}\]\[=\frac{46}{3}+\frac{23}{30}\]You need to login to perform this action.
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