Question

Suppose the events ${\mathbf{B}}_1\mathbf{,}{\mathbf{B}}_2\mathbf{,}{\mathbf{B}}_3$ are mutually exclusive and complementary events, such that$\mathbf{P}\mathbf{(}{\mathbf{B}}_1\mathbf{)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{2}$, $\mathbf{P}\mathbf{(}{\mathbf{B}}_2\mathbf{)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{4}$and $\mathbf{P}\mathbf{(}{\mathbf{B}}_3\mathbf{)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5}$. Consider another event A such that$\mathbf{P}\mathbf{\left(}\frac{\mathbf{A}}{{\mathbf{B}}_{\mathbf{1}}}\mathbf{\right)}\mathbf{=}\mathbf{P}\mathbf{\left(}\frac{\mathbf{A}}{{\mathbf{B}}_{\mathbf{2}}}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}$and$\mathbf{P}\mathbf{\left(}\frac{\mathbf{A}}{{\mathbf{B}}_{\mathbf{3}}}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{2}$Use Baye’s Rule to find

a.$\mathbf{P}\mathbf{\left(}\frac{{\mathbf{B}}_{\mathbf{1}}}{\mathbf{A}}\mathbf{\right)}$

b.$\mathbf{P}\left(\frac{{\mathbf{B}}_2}{A}\right)$

c.role="math" localid="1658214716845" $\mathbf{P}\mathbf{\left(}\frac{{\mathbf{B}}_{\mathbf{3}}}{\mathbf{A}}\mathbf{\right)}$

Short answer

Therefore, the values of all parts are:

1. 0.158
2. 0.073
3. 0.768

Step-by-step solution

Important formula

Baye’s formula is used for finding the conditional probability of the events.

The required formula is

$\begin{array}{c}\mathbf{P}\left(\frac{{\mathbf{B}}_i}{A}\right)\mathbf{=}\frac{\mathbf{P}\mathbf{(}{\mathbf{B}}_i\cap \mathbf{A}\mathbf{)}}{\mathbf{P}\mathbf{(}\mathbf{A})}\\ \mathbf{=}\frac{\mathbf{P}\mathbf{(}{\mathbf{B}}_i\mathbf{)}\mathbf{P}\mathbf{\left(}\frac{\mathbf{A}}{{\mathbf{B}}_{\mathbf{i}}}\mathbf{\right)}}{\mathbf{P}\mathbf{(}{\mathbf{B}}_1\mathbf{)}\mathbf{P}\mathbf{\left(}\frac{\mathbf{A}}{{\mathbf{B}}_{\mathbf{1}}}\mathbf{\right)}\mathbf{+}\mathbf{P}\mathbf{(}{\mathbf{B}}_2\mathbf{)}\mathbf{P}\mathbf{\left(}\frac{\mathbf{A}}{{\mathbf{B}}_{\mathbf{2}}}\mathbf{\right)}\mathbf{+}...\mathbf{+}\mathbf{P}\mathbf{(}{\mathbf{B}}_k\mathbf{)}\mathbf{P}\mathbf{\left(}\frac{\mathbf{A}}{{\mathbf{B}}_{\mathbf{k}}}\mathbf{\right)}}\end{array}$

(a) Find the value of  P(B1A)

Here $\mathbf{P}\mathbf{\left(}\frac{\mathbf{A}}{{\mathbf{B}}_{\mathbf{1}}}\mathbf{\right)}\mathbf{=}\mathbf{P}\mathbf{\left(}\frac{\mathbf{A}}{{\mathbf{B}}_{\mathbf{2}}}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}$, and$\mathbf{P}\mathbf{\left(}\frac{\mathbf{A}}{{\mathbf{B}}_{\mathbf{3}}}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{2}$.

Apply the baye’s formula, then

$\begin{array}{c}\mathbf{P}\left(\frac{{\mathbf{B}}_1}{A}\right)\mathbf{=}\frac{\mathbf{P}\mathbf{(}{\mathbf{B}}_1\mathbf{)}\mathbf{P}\mathbf{\left(}\frac{\mathbf{A}}{{\mathbf{B}}_{\mathbf{1}}}\mathbf{\right)}}{\mathbf{P}\mathbf{(}{\mathbf{B}}_1\mathbf{)}\mathbf{P}\left(\frac{A}{{\mathbf{B}}_1}\right)\mathbf{+}\mathbf{P}\mathbf{(}{\mathbf{B}}_2\mathbf{)}\mathbf{P}\mathbf{\left(}\frac{\mathbf{A}}{{\mathbf{B}}_{\mathbf{2}}}\mathbf{\right)}\mathbf{+}\mathbf{P}\mathbf{(}{\mathbf{B}}_3\mathbf{)}\mathbf{P}\mathbf{\left(}\frac{\mathbf{A}}{{\mathbf{B}}_{\mathbf{3}}}\mathbf{\right)}}\\ \mathbf{=}\frac{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{2}\mathbf{\left)}\mathbf{\right(}\mathbf{0}\mathbf{.}\mathbf{4}\mathbf{)}}{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{2}\mathbf{\left)}\mathbf{\right(}\mathbf{0}\mathbf{.}\mathbf{4}\mathbf{)}\mathbf{+}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{15}\mathbf{\left)}\mathbf{\right(}\mathbf{0}\mathbf{.}\mathbf{25}\mathbf{)}\mathbf{+}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{65}\mathbf{\left)}\mathbf{\right(}\mathbf{0}\mathbf{.}\mathbf{6}\mathbf{)}}\\ \mathbf{=}\mathbf{0}\mathbf{.}\mathbf{158}\end{array}$

So, the values of all parts are 0.158

(b) Evaluate the value of P(B2A)

$\begin{array}{c}\mathbf{P}\left(\frac{{\mathbf{B}}_2}{A}\right)\mathbf{=}\frac{\mathbf{P}\mathbf{(}{\mathbf{B}}_2\mathbf{)}\mathbf{P}\mathbf{\left(}\frac{\mathbf{A}}{{\mathbf{B}}_{\mathbf{2}}}\mathbf{\right)}}{\mathbf{P}\mathbf{(}{\mathbf{B}}_1\mathbf{)}\mathbf{P}\mathbf{\left(}\frac{\mathbf{A}}{{\mathbf{B}}_{\mathbf{1}}}\mathbf{\right)}\mathbf{+}\mathbf{P}\mathbf{(}{\mathbf{B}}_2\mathbf{)}\mathbf{P}\mathbf{\left(}\frac{\mathbf{A}}{{\mathbf{B}}_{\mathbf{2}}}\mathbf{\right)}\mathbf{+}\mathbf{P}\mathbf{(}{\mathbf{B}}_3\mathbf{)}\mathbf{P}\mathbf{\left(}\frac{\mathbf{A}}{{\mathbf{B}}_{\mathbf{3}}}\mathbf{\right)}}\\ \mathbf{=}\frac{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{15}\mathbf{\left)}\mathbf{\right(}\mathbf{0}\mathbf{.}\mathbf{25}\mathbf{)}}{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{2}\mathbf{\left)}\mathbf{\right(}\mathbf{0}\mathbf{.}\mathbf{4}\mathbf{)}\mathbf{+}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{15}\mathbf{\left)}\mathbf{\right(}\mathbf{0}\mathbf{.}\mathbf{25}\mathbf{)}\mathbf{+}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{65}\mathbf{\left)}\mathbf{\right(}\mathbf{0}\mathbf{.}\mathbf{6}\mathbf{)}}\\ \mathbf{=}\mathbf{0}\mathbf{.}\mathbf{073}\end{array}$

Hence, the values of all parts are 0.073

(c) Determine the value of P(B3A)

$\begin{array}{c}\mathbf{P}\left(\frac{{\mathbf{B}}_3}{A}\right)\mathbf{=}\frac{\mathbf{P}\mathbf{(}{\mathbf{B}}_3\mathbf{)}\mathbf{P}\mathbf{\left(}\frac{\mathbf{A}}{{\mathbf{B}}_{\mathbf{3}}}\mathbf{\right)}}{\mathbf{P}\mathbf{(}{\mathbf{B}}_1\mathbf{)}\mathbf{P}\left(\frac{A}{{\mathbf{B}}_1}\right)\mathbf{+}\mathbf{P}\mathbf{(}{\mathbf{B}}_2\mathbf{)}\mathbf{P}\left(\frac{A}{{\mathbf{B}}_2}\right)\mathbf{+}\mathbf{P}\mathbf{(}{\mathbf{B}}_3\mathbf{)}\mathbf{P}\mathbf{\left(}\frac{\mathbf{A}}{{\mathbf{B}}_{\mathbf{3}}}\mathbf{\right)}}\\ \mathbf{=}\frac{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{65}\mathbf{\left)}\mathbf{\right(}\mathbf{0}\mathbf{.}\mathbf{6}\mathbf{)}}{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{2}\mathbf{\left)}\mathbf{\right(}\mathbf{0}\mathbf{.}\mathbf{4}\mathbf{)}\mathbf{+}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{15}\mathbf{\left)}\mathbf{\right(}\mathbf{0}\mathbf{.}\mathbf{25}\mathbf{)}\mathbf{+}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{65}\mathbf{\left)}\mathbf{\right(}\mathbf{0}\mathbf{.}\mathbf{6}\mathbf{)}}\\ \mathbf{=}\mathbf{0}\mathbf{.}\mathbf{768}\end{array}$

Therefore, the values of all parts are 0.768