学校食堂每餐推出$A$、$B$两种套餐，某同学每天中午都会在食堂提供的两种套餐中选择一种套餐，若他前1天选择了$A$套餐，则第2天选择$A$套餐的概率为$\dfrac{1}{4}$；若他前1天选择了$B$套餐，则第2天选择了$A$套餐的概率为$\dfrac{3}{4}$.已知他开学第1天中午选择$A$套餐的概率为$\dfrac{2}{3}$，在该同学第3天选择了$A$套餐的条件下，他第2天选择$A$套餐的概率为______.
条件概率高频考法，贝叶斯公式应用全拆解！
$\dfrac{5}{26}$
设$A_n$为第$n$天选$A$套餐，$\overline{A_n}$为第$n$天选$B$套餐，
则$P(A_1) = \dfrac{2}{3}$，$P(\overline{A_1}) = 1 – \dfrac{2}{3} = \dfrac{1}{3}$，$P(A_{n+1}|A_n) = \dfrac{1}{4}$，$P(A_{n+1}|\overline{A_n}) = \dfrac{3}{4}$.
$\therefore P(A_2) = P(A_1)P(A_2|A_1) + P(\overline{A_1})P(A_2|\overline{A_1})$$= \dfrac{2}{3} \times \dfrac{1}{4} + \dfrac{1}{3} \times \dfrac{3}{4} = \dfrac{5}{12}$；
从而$P(\overline{A_2}) = 1 – \dfrac{5}{12} = \dfrac{7}{12}$，$P(A_3|A_2) = \dfrac{1}{4}$，$P(A_3|\overline{A_2}) = \dfrac{3}{4}$，
$P(A_3) = P(A_2)P(A_3|A_2) + P(\overline{A_2})P(A_3|\overline{A_2})$$= \dfrac{5}{12} \times \dfrac{1}{4} + \dfrac{7}{12} \times \dfrac{3}{4} = \dfrac{13}{24}$，
$P(A_2|A_3) = \dfrac{P(A_2)P(A_3|A_2)}{P(A_3)}$$= \dfrac{\dfrac{5}{12} \times \dfrac{1}{4}}{\dfrac{13}{24}} = \dfrac{\dfrac{5}{48}}{\dfrac{13}{24}} = \dfrac{5}{26}$.
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