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for two events, m and n, p(m) = 0.6, p(n|m) = 0.3, and p(n|m) = 0.3. fi…

Question

for two events, m and n, p(m) = 0.6, p(n|m) = 0.3, and p(n|m) = 0.3. find p(m|n).
p(m|n) = \frac{3}{5} (simplify your answer. type an integer or a fraction.)

Explanation:

Step1: Find $P(N\cap M)$

By the formula $P(N|M)=\frac{P(N\cap M)}{P(M)}$, we can get $P(N\cap M)=P(N|M)\times P(M)$. Substituting the given values, $P(N\cap M)=0.3\times0.6 = 0.18$.

Step2: Find $P(N\cap M')$

Since $P(M) = 0.6$, then $P(M')=1 - P(M)=1 - 0.6=0.4$. Using the formula $P(N|M')=\frac{P(N\cap M')}{P(M')}$, we have $P(N\cap M')=P(N|M')\times P(M')=0.3\times0.4 = 0.12$.

Step3: Find $P(N)$

By the law of total - probability, $P(N)=P(N\cap M)+P(N\cap M')$. So $P(N)=0.18 + 0.12=0.3$.

Step4: Find $P(M|N)$

By Bayes' theorem $P(M|N)=\frac{P(N\cap M)}{P(N)}$. Substituting the values of $P(N\cap M)$ and $P(N)$ we found above, $P(M|N)=\frac{0.18}{0.3}=\frac{3}{5}$.

Answer:

$\frac{3}{5}$