Q. 3.25 - A First Course in Probability

Question

Prove directly that,

$P(E\mid F)=P(E\mid FG)P(G\mid F)+P\left(E\mid F{G}^c\right)P\left(G^c\mid F\right).$

Step-by-Step Solution

Answer

By applying the definition of conditional probability,the direction starting from the right side.

Step: 1 Conditional probability:

The potential of an event or outcome occurring dependent on the existence of a preceding event or outcome is known as conditional probability. It's computed by multiplying the likelihood of the previous occurrence by the probability of the next, or conditional, event.

Step: 2 Proving equation:

By conditional probability,

$\begin{array}{l}P(E\mid FG)P(G\mid F)+P\left(E\mid F{G}^c\right)P\left(G^c\mid F\right)=\left(\frac{P\left(EFG\right)}{P\left(GF\right)}\times \frac{P\left(GF\right)}{P\left(F\right)}\right)+\left(\frac{P\left(EF{G}^c\right)}{P\left(G^{c}F\right)}\times \frac{P\left(G^{c}F\right)}{P\left(F\right)}\right)\\ P(E\mid FG)P(G\mid F)+P\left(E\mid F{G}^c\right)P\left(G^c\mid F\right)=\frac{P\left(EFG\right)}{P\left(F\right)}+\frac{P\left(EF{G}^c\right)}{P\left(F\right)}\\ P(E\mid FG)P(G\mid F)+P\left(E\mid F{G}^c\right)P\left(G^c\mid F\right)=\frac{P\left(EF\right)}{P\left(F\right)}\\ P(E\mid FG)P(G\mid F)+P\left(E\mid F{G}^c\right)P\left(G^c\mid F\right)=P(E\mid F).\end{array}$

Step: 3 Equating probability:

In total probability,

$$\begin{gathered} P_F=P(\cdot \mid F) \\ {P}_F(E\mid G)=\frac{P_F\left(EG\right)}{P_F\left(G\right)} \\ {P}_F(E\mid G)=\frac{P\left(EG\right[F)}{P(G\mid F)} \\ {P}_F(E\mid G)=\frac{\frac{P\left(EFG\right)}{P\left(F\right)}}{\frac{P\left(FG\right)}{P\left(F\right)}} \\ {P}_F(E\mid G)=P(E\mid FG). \end{gathered}$$