Question

for mutually exclusive events $r_1$, $r_2$, and $r_3$, we have $p(r_1)=0.05$, $p(r_2)=0.7$, and $p(r_3)=0.25$. also, $p(q|r_1)=0.4$, $p(q|r_2)=0.5$, and $p(q|r_3)=0.4$. find $p(r_3|q)$
$p(r_3|q)=square$
(type an integer or a simplified fraction.)

Explanation:

Step1: Apply the law of total - probability

The law of total - probability states that $P(Q)=\sum_{i = 1}^{3}P(R_i)P(Q|R_i)$.
$P(Q)=P(R_1)P(Q|R_1)+P(R_2)P(Q|R_2)+P(R_3)P(Q|R_3)$.
Substitute the given values: $P(R_1) = 0.05$, $P(Q|R_1)=0.4$, $P(R_2)=0.7$, $P(Q|R_2)=0.5$, $P(R_3)=0.25$, $P(Q|R_3)=0.4$.
$P(Q)=(0.05\times0.4)+(0.7\times0.5)+(0.25\times0.4)$.
$P(Q)=0.02 + 0.35+0.1$.
$P(Q)=0.47$.

Step2: Apply Bayes' theorem

Bayes' theorem states that $P(R_3|Q)=\frac{P(R_3)P(Q|R_3)}{P(Q)}$.
We know that $P(R_3)P(Q|R_3)=0.25\times0.4 = 0.1$ and $P(Q)=0.47$.
So, $P(R_3|Q)=\frac{0.1}{0.47}=\frac{10}{47}$.

Answer:

$\frac{10}{47}$