Giup minh voi

To solve this problem, we can use the concept of conditional probability. The overall probability can be calculated by considering all possible orders in which An, Bình, and Cường can be called to the board. We know that the first three students in the list are An, Bình, and Cường with probabilities 0.1, 0.2, and 0.3 respectively. The overall probability is then given by: \[ P(\text{An, Bình, Cường}) = P(\text{An}) \times P(\text{Bình}|\text{An}) \times P(\text{Cường}|\text{An,Bình}) \] Substitute the given probabilities: \[ P(\text{An, Bình, Cường}) = 0.1 \times 0.2 \times 0.3 = 0.006 \] Similarly, \[ P(\text{An, Cường, Bình}) = 0.1 \times 0.3 \times 0.2 = 0.006 \] \[ P(\text{Bình, An, Cường}) = 0.2 \times 0.1 \times 0.3 = 0.006 \] \[ P(\text{Bình, Cường, An}) = 0.2 \times 0.3 \times 0.1 = 0.006 \] \[ P(\text{Cường, An, Bình}) = 0.3 \times 0.1 \times 0.2 = 0.006\] \[ P(\text{Cường,Bình , An})=P(B,C,A)=P(C)P(B|C)P(A|B,C)=\frac {3}{10}\cdot\frac {2}{10}\cdot\frac {1}{10}=6\times {10}^{-4}=6\times {10}^{-4}.\] Finally adding all these probabilities gives us the overall probability: \[ P_{\text{overall}}=P(A,B,C)+P(A,C,B)+P(B,A,C)+P(B,C,A)+P(C,A,B)+P(C,B,A)=6\times {10}^{-4}+6\times {10}^{-4}+6\times {10}^{-4}+6\times {10}^{-4}+6\times {10}^{-4}=5/30=1/6=\\ =16.\overline {66666666666666}\% .\]