Suppose the events B1, B2, and B3 are mutually exclusive

Step 1 of 3

Let the events \(B_{1}, B_{2}\), and \(B_{3}\) are mutually exclusive and complementary events.

So \(P\left(B_{1}\right)=0.2, P\left(B_{2}\right)=0.15 \text {, }\) and \(P\left(B_{3}\right)=0.65\).

We consider another \(P\left(A / B_{1}\right)=0.4, P\left(A / B_{2}\right)=0.25\), and \(P\left(A / B_{3}\right)=0.6\).

Our goal is:

a). We need to find \(P\left(B_{1} / A\right)\).

b). We need to find \(P\left(B_{2} / A\right)\).

c). We need to find \(P\left(B_{3} / A\right)\).

a).

Now we use Bayes’s rule we have to find \(P\left(B_{1} / A\right)\).

First we need to find \(P\left(A \cap B_{1}\right)\).

\(P\left(A \cap B_{1}\right)=P\left(A / B_{1}\right) P\left(B_{1}\right)\)

We substitute \(P\left(A / B_{1}\right)\) and \(P\left(B_{1}\right)\) values.

\(\begin{array}{l}
P\left(A \cap B_{1}\right)=0.4 \times 0.2 \\
P\left(A \cap B_{1}\right)=0.8
\end{array}\)

Therefore, \(P\left(A \cap B_{1}\right)=0.8\).

Now we need to find \(P\left(A \cap B_{2}\right)\).

\(P\left(A \cap B_{2}\right)=P\left(A / B_{2}\right) P\left(B_{2}\right)\)

We substitute \(P\left(A / B_{2}\right)\) and \(P\left(B_{2}\right)\) values.

\(\begin{array}{l}
P\left(A \cap B_{2}\right)=0.25 \times 0.15 \\
P\left(A \cap B_{2}\right)=0.0375
\end{array}\)

Therefore, \(P\left(A \cap B_{2}\right)=0.0375\).

Then we need to find \(P\left(A \cap B_{3}\right)\).

\(P\left(A \cap B_{3}\right)=P\left(A / B_{3}\right) P\left(B_{3}\right)\)

We substitute \(P\left(A / B_{3}\right)\) and \(P\left(B_{3}\right)\) values.

\(\begin{array}{l}
P\left(A \cap B_{3}\right)=0.6 \times 0.65 \\
P\left(A \cap B_{3}\right)=0.39
\end{array}\)

Therefore, \(P\left(A \cap B_{3}\right)=0.39\)

Now we have to find P(A).

\(\mathrm{P}(\mathrm{A})=P\left(A \cap B_{1}\right)+P\left(A \cap B_{2}\right)+P\left(A \cap B_{3}\right)\)

We know that \(P\left(A \cap B_{1}\right), P\left(A \cap B_{2}\right)\), and \(P\left(A \cap B_{3}\right)\).

\(\begin{array}{l}
P(A)=0.8+0.0375+0.39 \\
P(A)=0.5075
\end{array}\)

Therefore, P(A) = 0.5075.

Now we use Bayes’s rule we have to find \(P\left(B_{1} / A\right)\).

The formula for \(P\left(B_{1} / A\right)\) is

\(P\left(B_{1} / A\right)=\frac{P\left(A \cap B_{1}\right)}{P(A)}\)

We know that \(P\left(A \cap B_{1}\right)=0.8\) and \(P(A)=0.5075\).

\(\begin{array}{l}
P\left(B_{1} / A\right)=\frac{0.8}{0.5075} \\
P\left(B_{1} / A\right)=0.642
\end{array}\)

Therefore, \(P\left(B_{1} / A\right)=0.642\).