hw4 110590049

tags probability

2023PB_HW4.pdf

1

$$\begin{gathered} P(s_1)=P(s_2)=\frac{1}{2} \\ P(r_1|s_1)=\frac{{\text{C}}_3^6}{{\text{C}}_3^{11}} \\ P(r_2|s_2)=\frac{{\text{C}}_3^9}{{\text{C}}_3^9} \\ \frac{P(r_1|s_1)P(s_1)}{P(r_1|s_1)P(s_1)+P(r_2|s_2)P(s_2)}=0.108 \end{gathered}$$

2

$$\begin{gathered} P(s_1)=\frac{{\text{C}}_1^5{\text{C}}_3^3}{{\text{C}}_4^8},P(s_2)=\frac{{\text{C}}_2^5{\text{C}}_2^3}{{\text{C}}_4^8},P(s_3)=\frac{{\text{C}}_3^5{\text{C}}_1^3}{{\text{C}}_4^8} \\ P(b_1|s_1)=\frac{{\text{C}}_1^3}{{\text{C}}_1^4},P(b_2|s_2)=\frac{{\text{C}}_1^2}{{\text{C}}_1^4},P(b_3|s_3)=\frac{{\text{C}}_1^1}{{\text{C}}_1^4} \\ \frac{P(b_2|s_2)P(s_2)}{P(b_1|s_1)P(s_1)+P(b_2|s_2)P(s_2)+P(b_3|s_3)P(s_3)}=\frac{4}{7}=0.571 \end{gathered}$$

3

$$\begin{gathered} \text{Yes.} \\ P(A)=\frac{18}{36} \\ P(B)=\frac{1}{6} \\ P(A|B)=P(A)=\frac{1}{2} \\ P(B|A)=P(B)=\frac{1}{6} \\ P(A)\times P(B)=P(A\cap B)=\frac{1}{12} \\ \text{A and B is independent event} \end{gathered}$$

4

$$\begin{gathered} P(\text{not hit})=(1-P(A))(1-P(B))(1-P(C))=0.006 \\ P(\text{hit})=1-P(\text{not hit})=0.994 \end{gathered}$$

5

$$\begin{gathered} P(\text{least once in four})=1-P(\text{not happen in four})=0.5904 \\ P(\text{not happen in four})=0.4096 \\ \text{since is independent trials so } \\ P(\text{occurrence in one trial})=1-P(\text{not happen in four}{)}^{\frac{1}{4}} \\ =1-0.8=0.2 \end{gathered}$$

6

$$1-\frac{5^6}{6^6}=0.665$$

7

$$\begin{gathered} P(E)=P(E|\text{head ace})+P(E|\text{head, not ace})+P(E|\text{not head ,ace})+P(E|\text{not head,not ace}) \\ P(E)=1\times \frac{1}{2}\frac{4}{52}+1\times \frac{1}{2}\frac{48}{52}+0\times \frac{1}{2}\frac{4}{52}+P(E)\times \frac{1}{2}\frac{48}{52} \\ P(E)=\frac{13}{14}=0.928 \end{gathered}$$