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Question: The probabilities that a student passes in Mathematics, Physics and Chemistry are *m,p* and *c*, re...

The probabilities that a student passes in Mathematics,

Physics and Chemistry are m,p and c, respectively. Of these

subjects, the student has 75% chance of passing in at least one, a 50% chance of passing in at least two, and a 40% chance of passing in exactly two. Which of the following

relations are true.

A

B

p+m+c=2720\mathrm { p } + \mathrm { m } + \mathrm { c } = \frac { 27 } { 20 }

C

pmc=110\mathrm { pmc } = \frac { 1 } { 10 }

D

pmc=14\mathrm { pmc } = \frac { 1 } { 4 }

Answer

p+m+c=2720\mathrm { p } + \mathrm { m } + \mathrm { c } = \frac { 27 } { 20 }

Explanation

Solution

We have, P(A∪B∪C)=34\mathrm { P } ( \mathrm { A } \cup \mathrm { B } \cup \mathrm { C } ) = \frac { 3 } { 4 }

i.e. P(A)+P(B)+P(C)−P(A∩B)\mathrm { P } ( \mathrm { A } ) + \mathrm { P } ( \mathrm { B } ) + \mathrm { P } ( \mathrm { C } ) - \mathrm { P } ( \mathrm { A } \cap \mathrm { B } ) −P(B∩C)−P(A∩C)+P(A∪B∪C)=34- \mathrm { P } ( \mathrm { B } \cap \mathrm { C } ) - \mathrm { P } ( \mathrm { A } \cap \mathrm { C } ) + \mathrm { P } ( \mathrm { A } \cup \mathrm { B } \cup \mathrm { C } ) = \frac { 3 } { 4 }

P(A∩B)+P(B∩C)+P(A∩C)−2P(A∩B∩C)=12\mathrm { P } ( \mathrm { A } \cap \mathrm { B } ) + \mathrm { P } ( \mathrm { B } \cap \mathrm { C } ) + \mathrm { P } ( \mathrm { A } \cap \mathrm { C } ) - 2 \mathrm { P } ( \mathrm { A } \cap \mathrm { B } \cap \mathrm { C } ) = \frac { 1 } { 2 } ……. (ii)

AndP(A∩B)+P(B∩C)+P(A∩C)−3P(A∩B∩C)=25\mathrm { P } ( \mathrm { A } \cap \mathrm { B } ) + \mathrm { P } ( \mathrm { B } \cap \mathrm { C } ) + \mathrm { P } ( \mathrm { A } \cap \mathrm { C } ) - 3 \mathrm { P } ( \mathrm { A } \cap \mathrm { B } \cap \mathrm { C } ) = \frac { 2 } { 5 }……. (iii)

From (ii) and (iii), we get

P(A∩B∩C)=12−25=110\mathrm { P } ( \mathrm { A } \cap \mathrm { B } \cap \mathrm { C } ) = \frac { 1 } { 2 } - \frac { 2 } { 5 } = \frac { 1 } { 10 } ……. (iv)

Ž P(A)P(B)P(C)=110⇒pmc=110\mathrm { P } ( \mathrm { A } ) \mathrm { P } ( \mathrm { B } ) \mathrm { P } ( \mathrm { C } ) = \frac { 1 } { 10 } \Rightarrow \mathrm { pmc } = \frac { 1 } { 10 }

From (i) , (ii) and (iii), we have

P(A)+P(B)+P(C)−(12+210)+110=34⇒p+m+c=2720\mathrm { P } ( \mathrm { A } ) + \mathrm { P } ( \mathrm { B } ) + \mathrm { P } ( \mathrm { C } ) - \left( \frac { 1 } { 2 } + \frac { 2 } { 10 } \right) + \frac { 1 } { 10 } = \frac { 3 } { 4 } \Rightarrow \mathrm { p } + \mathrm { m } + \mathrm { c } = \frac { 27 } { 20 }