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For three events \mathrm{A, B\: and\: C}, if \mathrm{P} (exactly one of the events \mathrm{A \: or\: B} occurs )\mathrm{=P} (exactly one of the events \mathrm{B\: or \: C} occurs) \mathrm{=P} (exactly one of the events \mathrm{C \: or \: A} occurs) \mathrm{=\rho \: and\: P} (all the three events occur simultaneously) \mathrm{=\rho^2}, where \mathrm{0<\rho<1 / 2}. Then, the probability of atleast one of the three events \mathrm{A, B \: and \: C} occurring, is
 

Option: 1

\mathrm{\frac{3 \rho+2 \rho^2}{2}}
 


Option: 2

\mathrm{\frac{\rho+3 \rho^2}{4}}
 


Option: 3

\mathrm{\frac{\rho+3 \rho^2}{2}}
 


Option: 4

\mathrm{\frac{3 \rho+2 \rho^2}{4}}


Answers (1)

We know that, \mathrm{P} (exactly one of \mathrm{A \: or\: B} occurs)

\mathrm{ =P(A)+P(B)-2 P(A \cap B) }

Therefore, \mathrm{ P(A)+P(B)-2 P(A \cap B)=\rho }.......(i)

Similarly, \mathrm{ P(B)+P(C)-2 P(B \cap C)=\rho }..........(ii)

and \mathrm{ \quad P(C)+P(A)-2 P(C \cap A)=\rho }...................(iii)

On adding Eqs. (i), (ii) and (iii), we get

\mathrm{ 2[P(A)+P(B)+P(C)-P(A \cap B) }\mathrm{ -P(B \cap C)-P(C \cap A)]=3 \rho }

\mathrm{ \Rightarrow P(A)+P(B)+P(C)-P(A \cap B) \\ -P(B \cap C)-P(C \cap A)=\frac{3 \rho}{2} \quad \ldots \text {(iv) } }
We are also given that

\mathrm{ P(A \cap B \cap C)=\rho^2 }...........(v)

Now, \mathrm{ P } (atleast one of \mathrm{ A, B\: and \: C } occurs)

\mathrm{ =P(A)+P(B)+P(C)-P(A \cap B)-P(B \cap C)-P(C \cap A) \mathrm{ +P(A \cap B \cap C) }}

\mathrm{ =\frac{3 \rho}{2}+\rho^2 } .......[from EqA. (iv) and (v)]

\mathrm{ =\frac{3 \rho+2 \rho^2}{2} }
Hence option 1 is correct.
 

Posted by

Ramraj Saini

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