hw6 1 2 3 4 5 6 7 8 9 10

hw6

tags probability

2023PB_HW6.pdf

1

p=212P(x=3)=Cx6×px×(1−p)6−x=0.0535p=\frac{2}{12}\\ P(x=3)=\text{C}_x^6\times p^x \times (1-p)^{6-x}=0.0535

2

p=0.45P(x)=Cx10×px×(1−p)10−xwhen x=4 the P(x) have maximum probability 0.2383p=0.45\\ P(x)=\text{C}_x^{10}\times p^x \times (1-p)^{10-x}\\ \text{when x=4 the P(x) have maximum probability } 0.2383

3

p=0.999P(x)=Cx8×p8−x×(1−p)xP(x={2,4,6,8})=0.0000278324891609p=0.999\\ P(x)=\text{C}_x^{8}\times p^{8-x} \times (1-p)^{x}\\ P(x=\{2,4,6,8\})=0.0000278324891609

4

p=0.025,E(x)=p∗80,λ=E(x)P(x)=e−λ×λxx!P(X>=2)=1−P(0)−P(1)=0.59399415029p=0.025,E(x)=p*80 ,\lambda=E(x)\\ P(x)=\frac{e^{-\lambda}\times \lambda^x}{x!}\\ P(X>=2)=1-P(0)-P(1)=0.59399415029

5

p=310,E(x)=p∗35,λ=E(X)P(x)=e−λ×λxx!P(x=10)=0.1236P(x=10)2=0.015278p=\frac{3}{10},E(x)=p*35 ,\lambda=E(X)\\ P(x)=\frac{e^{-\lambda}\times \lambda^x}{x!}\\ P(x=10)=0.1236\\ P(x=10)^2=0.015278

6

P(x)=e−λ×λxx!P(x=1)=e−λ×λ11!P(x=3)=e−λ×λ33!P(x=3)P(x=1)=λ23!=1λ=2.4494P(x=5)=0.0634449P(x)=\frac{e^{-\lambda}\times \lambda^x}{x!}\\ P(x=1)=\frac{e^{-\lambda}\times \lambda^1}{1!}\\ P(x=3)=\frac{e^{-\lambda}\times \lambda^3}{3!}\\ \frac{P(x=3)}{P(x=1)}=\frac{\lambda^2}{3!}=1\\ \lambda=2.4494\\ P(x=5)=0.0634449

7

p=0.2P(x,r)=Cr−1x−1×(1−p)x−r×prP(8,3)=0.05505p=0.2\\ P(x,r)=\text{C}_{r-1}^{x-1} \times (1-p)^{x-r}\times p^r\\ P(8,3)=0.05505

8

P(x)=px−1∗(1−p)P(x≥n)=1−∑i=1n−1P(i)P(x)=p^{x-1}*(1-p)\\ P(x\ge n) =1-\sum_{i=1}^{n-1}P(i)

9

p=0.15P(x)=C10−1x+10−1×p10×(1−p)xp=0.15\\ P(x)=\text{C}_{10-1}^{x+10-1}\times p^{10}\times(1-p)^{x}

10

p=50n,X∼Hgeo(n,50,50)d=50,t=50P(x)=CxdCt−xn−dCtnPn=624(x=4)Pn=625(x=4)≤1,Pn=626(x=4)Pn=625(x=4)≤1n=625p=\frac{50}{n}, X\sim Hgeo(n, 50, 50)\\ d=50,t=50\\ P(x)=\frac{\text{C}_x^d\text{C}_{t-x}^{n-d}}{\text{C}_t^n}\\ \frac{P_{n=624}(x=4)}{P_{n=625}(x=4)}\le1,\frac{P_{n=626}(x=4)}{P_{n=625}(x=4)}\le1\\\\ n=625