Q. 3.5 - A First Course in Probability

Question

(a) Prove that if $E$ and $F$ are mutually exclusive, then

$P(E\mid E\cup F)=\frac{P\left(E\right)}{P\left(E\right)+P\left(F\right)}$

(b) Prove that if $E_i,i\ge 1$ are mutually exclusive, then

$P\left(E_j\mid {\cup }_{i=1}^{\infty }{E}_i\right)=\frac{P\left(E_j\right)}{{\sum }_{i=1}^{\infty }P\left(E_i\right)}$

Step-by-Step Solution

Answer

We concluded that 

(a) If $E$ and $F$ are mutually exclusive then $P(E\mid E\cup F)=\frac{P\left(E\right)}{P\left(E\right)+P\left(F\right)}$

(b) If $E_i,i\ge 1$ are mutually exclusive then $P\left(Ej\mid {\cup }_{i=1}^{\infty }{E}_i\right)=\frac{P\left(E_j\right)}{{\sum }_{i=1}^{\infty }P\left(E_i\right)}$.

Step 1: Concept Introduction Part(a)

Mutually exclusive is a statistical word defining two or more possibilities that cannot occur simultaneously. It is normally used to represent a case where the happening of one output replaces the other.

Step 2: Explanation Part(a)

If $E$ and $F$ are mutally exclusive then $E\cap F=\varnothing$.

 Since $E\cap (E\cup F)=E$ and by aditivity $P(E\cup F)=P\left(E\right)+P\left(F\right)$,

we conclude that

$P(E\mid E\cup F)=\frac{P(E\cap (E\cup F\left)\right)}{P(E\cup F)}$

$=\frac{P\left(E\right)}{P\left(E\right)+P\left(F\right)}$.

Step 3: Final Answer Part(a)

$P(E\mid E\cup F)=\frac{P(E\cap (E\cup F\left)\right)}{P(E\cup F)}=\frac{P\left(E\right)}{P\left(E\right)+P\left(F\right)}$

Step 4: Concept Introduction Part(b)

Mutually exclusive is a statistical word defining two or more possibilities that cannot occur simultaneously. It is normally used to represent a case where the happening of one output replaces the other.

Step 5: Explanation Part (b)

Since

$E_j\cap \underset{i=1}{\overset{\infty }{\cup }}{E}_i=\underset{i=1}{\overset{\infty }{\cup }}\left(E_j\cap E_i\right)=\left[i\ne j⟹E_j\cap E_i=\varnothing \right]=E_j$

$P\left(\underset{i=1}{\overset{\infty }{\cup }}{E}_i\right)=[\sigma -\text{ aditivity }/\text{ Axiom }3]=\sum _{i=1}^{\infty }P\left(E_i\right)$

We concluded that 

$P\left(E_j\mid \underset{i=1}{\overset{\infty }{\cup }}{E}_i\right)=\frac{P\left(E_j\cap {\cup }_{i=1}^{\infty }{E}_i\right)}{P\left({\cup }_{i=1}^{\infty }{E}_i\right)}$

$=\frac{P\left(E_j\right)}{{\sum }_{i=1}^{\infty }P\left(E_i\right)}$.

Step 6: Final Answer Part(b)

$P\left(E_j\mid \underset{i=1}{\overset{\infty }{\cup }}{E}_i\right)=\frac{P\left(E_j\cap {\cup }_{i=1}^{\infty }{E}_i\right)}{P\left({\cup }_{i=1}^{\infty }{E}_i\right)}=\frac{P\left(E_j\right)}{{\sum }_{i=1}^{\infty }P\left(E_i\right)}$