Question

the prior probabilities for events $a_1$ and $a_2$ are $p(a_1)=0.50$ and $p(a_2)=0.50$. it is also known that $p(a_1cap a_2)=0$. suppose $p(b|a_1)=0.20$ and $p(b|a_2)=0.40$. a. are events $a_1$ and $a_2$ mutually exclusive? explain. (i) $p(a_1cap a_2)=0$ (ii) $p(a_1)+p(a_2)=1$ (iii) $p(a_2)
eq p(a_2|a_1)$ (iv) $p(a_2)
eq p(a_2|a_1)$ choice (i) b. compute $p(a_1cap b)$ (to 2 decimals). compute $p(a_2cap b)$ (to 2 decimals). c. compute $p(b)$ (to 2 decimals). d. apply bayes theorem to compute $p(a_1|b)$ (to 4 decimals).

Explanation:

Step1: Recall the formula for conditional - probability

The formula for conditional probability is $P(B|A)=\frac{P(A\cap B)}{P(A)}$, which can be rewritten as $P(A\cap B)=P(B|A)P(A)$.

Step2: Calculate $P(A_1\cap B)$

Given $P(B|A_1) = 0.20$ and $P(A_1)=0.50$. Using the formula $P(A_1\cap B)=P(B|A_1)P(A_1)$, we have $P(A_1\cap B)=0.20\times0.50 = 0.10$.

Step3: Calculate $P(A_2\cap B)$

Given $P(B|A_2)=0.40$ and $P(A_2)=0.50$. Using the formula $P(A_2\cap B)=P(B|A_2)P(A_2)$, we get $P(A_2\cap B)=0.40\times0.50 = 0.20$.

Step4: Use the law of total - probability

Since $A_1$ and $A_2$ are mutually exclusive, $P(B)=P(A_1\cap B)+P(A_2\cap B)$. Substituting the values of $P(A_1\cap B)$ and $P(A_2\cap B)$ we found above, $P(B)=0.10 + 0.20=0.30$.

Step5: Apply Bayes' theorem

Bayes' theorem states that $P(A_i|B)=\frac{P(B|A_i)P(A_i)}{P(B)}$. For $i = 1$, $P(A_1|B)=\frac{P(B|A_1)P(A_1)}{P(B)}=\frac{0.20\times0.50}{0.30}=\frac{0.10}{0.30}\approx0.3333$. For $i = 2$, $P(A_2|B)=\frac{P(B|A_2)P(A_2)}{P(B)}=\frac{0.40\times0.50}{0.30}=\frac{0.20}{0.30}\approx0.6667$.

Answer:

b. $P(A_1\cap B)=0.10$, $P(A_2\cap B)=0.20$
c. $P(B)=0.30$
d. $P(A_1|B)\approx0.3333$, $P(A_2|B)\approx0.6667$