Verify Equation 1.2.P(a1<Xa2,b1<Yb2)=F(a2,b2)+F(a1,b1)F(a1,b2)F(a2,b1)

Short Answer

Expert verified

Equation is proved.

P(a1<Xa2,b1<Yb2)=F(a2,b2)+F(a1,b1)F(a1,b2)F(a2,b1)

Step by step solution

01

Equation : 

A declaration (represented by the symbol =) that the values of two mathematical expressions are equal.

02

Explanation :  

Let us define the events,

A:Xa1B:Xa2C:Yb1D:Yb2

Pa1<X<a2b1Yb2=PB-A(D-C)=PBD-C-AD-C......(1)[Bydistributivelaw]

We know that if EFEF=E, then

P(F-E)=P(E¯F)=P(F)-P(EF)=P(F)-P(E).......(2)

Obviously ABAD-CBD-C

Hence by using (2), we get from (1)

Pa1<Xa2b1<Yb2=PBD-C-PAD-C=PBD-BC-PAD-AC=PBD-BC-PAD+PAC......(3)

We have localid="1647341297556" P(BD)=PXa2Yb2=Fa2,b2

Similarly localid="1647341325673" P(BC)=F(a2,b1)P(AD)=F(a1,b2)

Substitute these values in (3)we get

P(a1<Xa2,b1<Yb2)=F(a2,b2)+F(a1,b1)F(a1,b2)F(a2,b1)

Hence proved.

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