Question

for mutually exclusive events $r_1$, $r_2$, and $r_3$, we have $p(r_1)=0.05$, $p(r_2)=0.5$, and $p(r_3)=0.45$. also, $p(q|r_1)=0.3$, $p(q|r_2)=0.4$, and $p(q|r_3)=0.2$. find $p(r_1|q)$.
$p(r_1|q)=square$ (type an integer or a simplified fraction.)

Explanation:

Step1: Recall Bayes' theorem

$P(R_1|Q)=\frac{P(Q|R_1)P(R_1)}{P(Q)}$

Step2: Calculate $P(Q)$ using the law of total - probability

$P(Q)=P(Q|R_1)P(R_1)+P(Q|R_2)P(R_2)+P(Q|R_3)P(R_3)$
$P(Q)=(0.3\times0.05)+(0.4\times0.5)+(0.2\times0.45)$
$P(Q)=0.3\times0.05 + 0.4\times0.5+0.2\times0.45=0.015 + 0.2+0.09=0.305$

Step3: Calculate $P(R_1|Q)$

$P(R_1|Q)=\frac{P(Q|R_1)P(R_1)}{P(Q)}=\frac{0.3\times0.05}{0.305}=\frac{0.015}{0.305}=\frac{15}{305}=\frac{3}{61}$

Answer:

$\frac{3}{61}$