91Ó°ÊÓ

Events:

G - a person is in the good risk category

M-a person is in the average risk category

B - a person is in the bad risk category

A - a person has an accident in a fixed year

Probabilities:$P\left(G\right)=20\%=0.2P\left(M\right)=50\%=0.5P\left(B\right)=30\%=0.3$

Since the boxes' contents are known:

$P(A\backslash G)=0.05P(A\backslash M)=0.15P(A\backslash B)=0.30$

Calculate: $\left.P\left(A\right),P\left(G∪Mâˆ\pounds A^c\right)\right\}$

P(A) can be obtained by conditioning on whether a person is a good, an average or a bad risk:

$P\left(A\right)=P(Aâˆ\pounds G)P\left(G\right)+P(Aâˆ\pounds M)P\left(M\right)+P(Aâˆ\pounds B)P\left(B\right)$

By substituting before stated probabilities the result is :

$P\left(A\right)=0.175$

Conditional probability is also a probability ,and G and M are mutually exclusive, so by Axiom 3:

$P\left(G∪Mâˆ\pounds A^c\right)=P\left(Gâˆ\pounds A^c\right)+P\left(Mâˆ\pounds A^c\right)$

For $P\left(Gâˆ\pounds A^c\right)$use the definition formula for conditional probability:

$P\left(Gâˆ\pounds A^c\right)=\frac{P\left(G{A}^c\right)}{P\left(A^c\right)}$

As$P\left(A\right)=0.175$is already calculated, a proposition states :

$P\left(A^c\right)=1-P\left(A\right)⇒P\left(A^c\right)=0.825$

And by a different use of the definition of $P\left(A^câˆ\pounds G\right)$:

$P\left(G{A}^C\right)=P\left(A^C/G\right)P\left(G\right)$

Conditional probability is also a probability so:

$P\left(A^C/G\right)=1-P\left(A/G\right)⇒P\left(A^C/G\right)=0.95$

Now all the elements of the equation are known, so we can calculate :

$P\left(G/A^C\right)=\frac{0.95×0.2}{0.825}≈0.2303$

The same is applied for $P\left(M/A^C\right)$:

$P\left(M/A^C\right)=\frac{P\left(M{A}^C\right)}{P\left(\left(A^C\right)\right)}$

$P\left(A^{C^{}}\right)=0.825$

$$\begin{gathered} P\left(M{A}^C\right)=P\left(A^C/M\right)P\left(M\right),P\left(A^{C}M\right)=1-P\left(A/M\right) \\ ⇒P\left(M{A}^C\right)=0.85×0.5=0.425 \\ P\left(G/A^C\right)=\frac{0.425}{0.825}≈0.5152 \end{gathered}$$

Conditional probability is also a probability so:

$P\left(G∪M/A^C\right)≈0.7455$