hw4 1 2 3 4 5 6 7

hw4

tags probability

2023PB_HW4.pdf

1

P(s1)=P(s2)=12P(r1∣s1)=C36C311P(r2∣s2)=C39C39P(r1∣s1)P(s1)P(r1∣s1)P(s1)+P(r2∣s2)P(s2)=0.108P(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

2

P(s1)=C15C33C48,P(s2)=C25C23C48,P(s3)=C35C13C48P(b1∣s1)=C13C14,P(b2∣s2)=C12C14,P(b3∣s3)=C11C14P(b2∣s2)P(s2)P(b1∣s1)P(s1)+P(b2∣s2)P(s2)+P(b3∣s3)P(s3)=47=0.571P(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

3

Yes.P(A)=1836P(B)=16P(A∣B)=P(A)=12P(B∣A)=P(B)=16P(A)×P(B)=P(A∩B)=112A and B is independent event \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}

4

P(not hit)=(1−P(A))(1−P(B))(1−P(C))=0.006P(hit)=1−P(not hit)=0.994P(\text{not hit})=(1-P(A))(1-P(B))(1-P(C))=0.006\\ P(\text{hit})=1-P(\text{not hit})=0.994

5

P(least once in four)=1−P(not happen in four)=0.5904P(not happen in four)=0.4096since is independent trials so P(occurrence in one trial)=1−P(not happen in four)14=1−0.8=0.2P(\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

6

1−5666=0.6651-\frac{5^6}{6^6}=0.665

7

P(E)=P(E∣head ace)+P(E∣head, not ace)+P(E∣not head ,ace)+P(E∣not head,not ace)P(E)=1×12452+1×124852+0×12452+P(E)×124852P(E)=1314=0.928P(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