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Question

If P(A¯B)+P(AB)=1-k, P(AC)+P(AC)=1-2K, P(BC)+P(CB)=1-K and P(ABC)=K2 then find the value of P(at least one of A, B, C) is


A

>12

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B

18,12

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C

<14

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D

14

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Solution

The correct option is A

>12


Step-1: Apply the formula of Sets:

Given P(A¯B)+P(AB)=1-k, --------(i)

P(AC)+P(AC)=1-2K, --------(ii)

P(BC)+P(CB)=1-K ----------(iii)

and P(ABC)=K2 ---------(iv)

By the formula of Sets, we have

P(AB)=P(B)-P(AB)

Apply the above formula to L.H.S of the equations (i)

P(A¯B)+P(AB)=1-kP(B)-P(AB)+P(A)-P(AB)=1-kP(A)+P(B)-2P(AB)=1-K

From the above equation, we can write

P(AB)=P(A)+P(B)-1+k2 ….(v)

Apply the above formula to L.H.S of the equations (iI)

P(A¯C)+P(AC)=1-2kP(C)-P(AC)+P(A)-P(AC)=1-2kP(A)+P(C)-2P(AC)=1-2K

From the above equation, we can write

P(AC)=P(A)+P(C)-1+2k2 …(vi)

Apply the above formula to L.H.S of the equations (iii)

P(B¯C)+P(BC)=1-kP(C)-P(BC)+P(B)-P(BC)=1-kP(B)+P(C)-2P(BC)=1-k

From the above equation, we can write

P(BC)=P(B)+P(C)-1+k2 ….(vii)

Step 2: Find the value of P(ABC)

We also have the formula as

P(ABC)=P(A)+P(B)+P(C)-P(AB)-P(BC)-P(AC)+P(ABC)

Now substitute the values of (iv), (v), (vi),(vii) in the above equation

P(ABC)=P(A)+P(B)-P(C)-P(AB)-P(BC)-P(AC)+P(ABC)=P(A)+P(B)-P(C)-P(A)+P(B)-1+k2-P(B)+P(C)-1+k2-P(A)+P(C)-1+2k2+k2=2k2-4k+32

Step 3: Apply P(ABC)=P(atleastoneofA,B,C)

Since,

P(ABC)=2k2-4k+32

P(atleastoneofA,B,C)= 2k2-4k+32

But 2k2-4k+32>12

Therefore, P(atleastoneofA,B,C)> 12

Hence, option(A) is correct.


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