Solve the problem using Bayes' Theorem. Round the answer to the nearest hundredth, if necessary.
-For mutually exclusive events $X_{1}, X_{2}$, and $X_{3}$, let $P\left(X_{1}\right)=.15, P\left(X_{2}\right)=.64$, and $P\left(X_{3}\right)=.21$. Also, $\mathrm{P}\left(\mathrm{Y} \mid \mathrm{X}_{1}\right)=.40, \mathrm{P}\left(\mathrm{Y} \mid \mathrm{X}_{2}\right)=.30$, and $\mathrm{P}\left(\mathrm{Y} \mid \mathrm{X}_{3}\right)=.60$. Find $\mathrm{P}\left(\mathrm{X}_{1} \mid \mathrm{Y}\right)$.
A) .33
B) .12
C) .16
D) .51

Question

Quizplus · Accepted Answer

Using Bayes' Theorem: $ P(X_1 \mid Y) = \frac{P(Y \mid X_1) \cdot P(X_1)}{P(Y)} $. First, calculate $ P(Y) = P(Y \mid X_1) \cdot P(X_1) + P(Y \mid X_2) \cdot P(X_2) + P(Y \mid X_3) \cdot P(X_3) = 0.40 \cdot 0.15 + 0.30 \cdot 0.64 + 0.60 \cdot 0.21 = 0.06 + 0.192 + 0.126 = 0.378 $. Then, $ P(X_1 \mid Y) = \frac{0.06}{0.378} \approx 0.16 $.

Solve the Problem Using Bayes' Theorem X1,X2X_{1}, X_{2}X1​,X2​ , And X3X_{3}X3​ , Let

Solve the Problem Using Bayes' Theorem $X_{1}, X_{2}$ , And $X_{3}$ , Let