A) \[p+m+c=\frac{19}{20}\]
B) \[p+m+c=\frac{27}{20}\]
C) \[pmc=\frac{1}{10}\]
D) \[pmc=\frac{1}{4}\]
Correct Answer: B
Solution :
Let \[M,\,\,P\] and \[C\] be the events of passing in mathematics, physics and chemistry respectively. \[P(M\cup P\cup C)=\frac{75}{100}=\frac{3}{4}\] \[P(M\cap P)+P(P\cap C)+P(M\cap C)-2P(M\cap P\cap C)=\frac{50}{100}=\frac{1}{2}\] \[P(M\cap P)+P(P\cap C)+P(M\cap C)-2P(M\cap P\cap C)=\frac{40}{100}=\frac{2}{5}\] \ \[m(1-p)(1-c)+p(1-m)(1-c)+c(1-m)(1-p)\] \[+mp(1-c)+mc(1-p)+pc(1-m)+mpc=\frac{3}{4}\] \[\Rightarrow m+p+c-mc-mp-pc+mpc=\frac{3}{4}\] .....(i) Similarly, \[mp(1-c)+pc(1-m)+mc(1-p)+mpc=\frac{1}{2}\] \[\Rightarrow mp+pc+mc-2mpc=\frac{1}{2}\] .....(ii) \[mp(1-c)+pc(1-m)+mc(1-p)=\frac{2}{5}\] \[\Rightarrow mp+pc+mc-3mpc=\frac{2}{5}\] .....(iii) From (ii) to (iii), \[mpc=\frac{1}{2}-\frac{2}{5}=\frac{1}{10}\] From (i) and (ii), \[m+p+c-mpc=\frac{3}{4}+\frac{1}{2}\] \[\therefore \,\,\,m+p+c=\frac{3}{4}+\frac{1}{2}+\frac{1}{10}=\frac{15+10+2}{20}=\frac{27}{20}.\]
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