Question

Suppose the events B1, ${\mathbf{B}}_{\mathbf{1}}\mathbf{,}{\mathbf{B}}_{\mathbf{2}}\mathbf{,}{\mathbf{B}}_{\mathbf{3}}$are mutually exclusive and complementary events, such that, $\mathbf{P}\mathbf{(}{\mathbf{B}}_{\mathbf{1}}\mathbf{)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{2}$, $\mathbf{P}\mathbf{(}{\mathbf{B}}_{\mathbf{2}}\mathbf{)}\mathbf{=}\mathbf{0}.15$and $\mathbf{P}\mathbf{(}{\mathbf{B}}_{\mathbf{3}}\mathbf{)}\mathbf{=}\mathbf{0}.6\mathbf{5}$. Consider another event A such that $\mathbf{P}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{4}$. If A is independent ${\mathbf{B}}_{\mathbf{1}}\mathbf{,}{\mathbf{B}}_{\mathbf{2}}\mathbf{,}{\mathbf{B}}_{\mathbf{3}}$, use Bayes’s Rule to show that$\mathbf{P}\left(\frac{{\mathbf{B}}_{\mathbf{1}}}{\mathbf{A}}\right)\mathbf{=}\mathbf{P}\mathbf{(}{\mathbf{B}}_{\mathbf{1}}\mathbf{)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{2}$

Short answer

Applying the Byes’s rule I get the result$\mathrm{P}\left(\frac{{\mathrm{B}}_1}{\mathrm{A}}\right)=\mathrm{P}({\mathrm{B}}_1)=0.2$

Step-by-step solution

Important formula

The Baye’s formula is

$\begin{array}{rcl}\mathbf{P}\left(\frac{{\mathbf{B}}_{\mathbf{i}}}{\mathbf{A}}\right)& \mathbf{=}& \frac{\mathbf{P}\mathbf{(}{\mathbf{B}}_{\mathbf{i}}\cap \mathbf{A}\mathbf{)}}{\mathbf{P}\mathbf{(}\mathbf{A})}\\ & \mathbf{=}& \frac{\mathbf{P}\mathbf{(}{\mathbf{B}}_{\mathbf{i}}\mathbf{)}\mathbf{P}\left(\frac{\mathbf{A}}{{\mathbf{B}}_{\mathbf{i}}}\right)}{\mathbf{P}\mathbf{(}{\mathbf{B}}_{\mathbf{1}}\mathbf{)}\mathbf{P}\left(\frac{\mathbf{A}}{{\mathbf{B}}_{\mathbf{1}}}\right)\mathbf{+}\mathbf{P}\mathbf{(}{\mathbf{B}}_{\mathbf{2}}\mathbf{)}\mathbf{P}\left(\frac{\mathbf{A}}{{\mathbf{B}}_{\mathbf{2}}}\right)\mathbf{+}...\mathbf{+}\mathbf{P}\mathbf{(}{\mathbf{B}}_{\mathbf{k}}\mathbf{)}\mathbf{P}\left(\frac{\mathbf{A}}{{\mathbf{B}}_{\mathbf{k}}}\right)}\\ & & \end{array}$

Show that PB1A=P(B1)=0.2

Apply the Baye’s rule which says that there are mutually exclusive and exhaustive$B_1,B_2,B_3,...,B_k$ such that $P\left(B_1\right)+P\left(B_2\right)+...P\left(B_k\right)=1$and another event A then .

$P\left(\frac{B_i}{A}\right)=\frac{P\left(B_i\cap A\right)}{P\left(A\right)}$

Now,

$\begin{array}{rcl}P\left(B_1\right)+P\left(B_2\right)+P\left(B_3\right)& =& 0.2+0.15+0.65\\ & =& 1\\ & & \end{array}$

A does not depends upon$B_1,B_2,B_3$

So,

$\begin{array}{rcl}P\left(\frac{B_1}{A}\right)& =& \frac{P\left(B_1\cap A\right)}{P\left(A\right)}\\ & =& \frac{P\left(B_1\right).P\left(A\right)}{P\left(A\right)}\\ & =& P\left(B_1\right)\\ & =& 0.2\\ & & \end{array}$

Therefore, this shows that$P\left(\frac{B_1}{A}\right)=P\left(B_1\right)=0.2$