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 $P(x_1)=0\mathrm{.}31,P(x_2)=0\mathrm{.}17,$ and $P(x_3)=0\mathrm{.}52.$ Also, $P(Y|x_1)=0\mathrm{.}40,P(Y|x_2)=0\mathrm{.}30,$ and $P(Y|x_3)=0\mathrm{.}60.$ Find $P(x_1|Y).$

A$.0\mathrm{.}10$

B$.0\mathrm{.}64$

C$.0\mathrm{.}19$

D$.0\mathrm{.}25$