Use Bayes' rule
Suppose "B_1,B_2,..., B_n" are mutually exclusive and exhaustive cases (hypotheses) and "A" is some event (evidence) that has been observed. Then
"P(B_i|A)={P(B_i)P(A|B_i) \\over \\displaystyle\\sum_{j=1}^nP(B_j)P(A|B_j)}, 1\\leq i\\leq n"
"P(B_1|A)={P(B_1)P(A|B_1) \\over P(B_1)P(A|B_1)+P(B_2)P(A|B_2)+P(B_3)P(A|B_3)}"
"P(B_2|A)={P(B_2)P(A|B_2) \\over P(B_1)P(A|B_1)+P(B_2)P(A|B_2)+P(B_3)P(A|B_3)}"
"P(B_3|A)={P(B_3)P(A|B_3) \\over P(B_1)P(A|B_1)+P(B_2)P(A|B_2)+P(B_3)P(A|B_3)}"
Given that
"P(B_1)=P(B_2)=P(B_3)""P(A|B_1)=0.75,P(A|B_2)=0.84,P(A|B_3)=0.93"
We assume that "P(B_1)=P(B_2)=P(B_3)\\not=0". We don't need to compute "P(B_1),P(B_2),P(B_3)"
"P(B_1|A)={0.75\\cdot P(B_1)\\over 0.75\\cdot P(B_1)+0.84\\cdot P(B_2)+0.93\\cdot P(B_3)}="
"={0.75\\cdot P(B_1)\\over 0.75\\cdot P(B_1)+0.84\\cdot P(B_1)+0.93\\cdot P(B_1)}="
"={0.75\\over 0.75+0.84+0.93}={25 \\over 84}"
"P(B_2|A)={0.84\\cdot P(B_2)\\over 0.75\\cdot P(B_1)+0.84\\cdot P(B_2)+0.93\\cdot P(B_3)}="
"={0.84\\cdot P(B_2)\\over 0.75\\cdot P(B_2)+0.84\\cdot P(B_2)+0.93\\cdot P(B_2)}="
"={0.84\\over 0.75+0.84+0.93}={28 \\over 84}"
"P(B_3|A)={0.93\\cdot P(B_3)\\over 0.75\\cdot P(B_1)+0.84\\cdot P(B_2)+0.93\\cdot P(B_3)}="
"={0.93\\cdot P(B_3)\\over 0.75\\cdot P(B_3)+0.84\\cdot P(B_3)+0.93\\cdot P(B_3)}="
"={0.93\\over 0.75+0.84+0.93}={31 \\over 84}"
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