Solve the problem using Bayes' Theorem. Round the answer to the nearest hundredth, if necessary.
-For two events $M$ and $N, P(M)=.4, P(N \mid M)=.6$, and $P\left(N \mid M^{\prime}\right)=.2$. Find $P\left(M^{\prime} \mid N\right)$.
A) .33
B) .67
C) 0
D) 1.0

Question

Quizplus · Accepted Answer

Using Bayes' Theorem, $P(M' \mid N) = \frac{P(N \mid M')P(M')}{P(N)}$. First, find $P(N)$ using the law of total probability: $P(N) = P(N \mid M)P(M) + P(N \mid M')P(M') = 0.6 \times 0.4 + 0.2 \times 0.6 = 0.24 + 0.12 = 0.36$. Then, $P(M') = 1 - P(M) = 0.6$. So, $P(M' \mid N) = \frac{0.2 \times 0.6}{0.36} = \frac{0.12}{0.36} = 0.33$.

Solve the Problem Using Bayes' Theorem MMM And N,P(M)=.4,P(N∣M)=.6N, P(M)=.4, P(N \mid M)=.6N,P(M)=.4,P(N∣M)=.6

Solve the Problem Using Bayes' Theorem $M$ And $N, P(M)=.4, P(N \mid M)=.6$