5. Th e probability of snow on any given day in a particular region of the North

Question

5. Th e probability of snow on any given day in a particular region of the North Pole is .90. Suppose you wish to simulate whether or not it snows on 3 consecutive days. You perform 10 trials, where cach trial consists of three randomly generated numbers from 1 to 10. a.) Explain how you will use the digits 1-10 to simulate the probability of it snowing or not on each of the three consecutive days. (b) Clearly list the results from all of your 10 trials. (c) Based on your simulation, what is the probability (estimated) that it will on exactly one day? How about exactly two of the days? snow

Explanation / Answer

Probability of snow on any given day=0.9.

Now we want to perform 10 trials where each trial consists of three randomly generated numbers from 1 to 10:

1) 7, 5, 10

2) 6, 8, 6

3) 8, 8, 9

4) 3, 9, 7

5) 1, 8, 6

6) 5, 9, 8

7) 7, 4, 10

8) 7, 6, 8

9) 1, 4, 8

10) 8, 3, 3

(a) &(b) Now we divide each random number by 10 of each trial and then multiply these three then we get simulated probability of snowing three consecutive days. The calculations are shown below:

Sl. no. Random no.s simulated probabilities

after divided by 10

1) 0.7, 0.5, 0.1   0.035

2) 0.6, 0.8, 0.6   0.288

3) 0.8, 0.8, 0.9   0.576

4) 0.3, 0.9, 0.7   0.189

5) 0.1, 0.8, 0.6   0.048

6) 0.5, 0.9, 0.8    0.36

7) 0.7, 0.4, 0.1 0.028

8) 0.7, 0.6, 0.8 0.336

9) 0.1, 0.4, 0.8   0.032

10) 0.8, 0.3, 0.3 0.072

The simulated probability that it snows thee consecutive days=(0.035+0.288+0.576+0.189+0.048+0.36+0.028+

0.336+0.032+0.072)/10=0.1964

(c) Followings are the simulated probabilities of snowing exactly one day:

1. (1-0.7)*(1-0.5)*0.1+0.7*(1-0.5)*(1-0.1)+(1-0.7)*0.5*(1-0.1)=0.465

2.(1-0.6)*(1-0.8)*0.6+0.6*(1-0.8)*(1-0.6)+(1-0.6)*0.8*(1-0.6)=0.224

3.(1-0.8)*(1-0.8)*0.9+0.8*(1-0.8)*(1-0.9)+(1-0.8)*0.8*(1-0.9)=0.068

4.(1-0.3)*(1-0.9)*0.7+0.3*(1-0.9)*(1-0.7)+(1-0.3)*0.9*(1-0.7)=0.247

5.(1-0.1)*(1-0.8)*0.6+0.1*(1-0.8)*(1-0.6)+(1-0.1)*0.8*(1-0.6)=0.404

6.(1-0.5)*(1-0.9)*0.8+0.5*(1-0.9)*(1-0.8)+(1-0.5)*0.9*(1-0.8)=0.14

7.(1-0.7)*(1-0.4)*0.1+0.7*(1-0.4)*(1-0.1)+(1-0.7)*0.4*(1-0.1)=0.504

8.(1-0.7)*(1-0.6)*0.8+0.7*(1-0.6)*(1-0.8)+(1-0.7)*0.6*(1-0.8)=0.188

9.(1-0.1)*(1-0.4)*0.8+0.1*(1-0.4)*(1-0.8)+(1-0.1)*0.4*(1-0.8)=0.516

10.(1-0.8)*(1-0.3)*0.3+0.8*(1-0.3)*(1-0.3)+(1-0.8)*0.3*(1-0.3)=0.476

Estimated probability of snowing exactly one day=

(0.465+0.224+0.068+0.247+0.404+0.14+0.504+0.188+0.516+0.476)/10=0.3232

Followings are the simulated probabilities of snowing exactly two days:

1. (1-0.7)*0.5*0.1+0.7*(1-0.5)*0.1+0.7*0.5*(1-0.1) =0.365

2. (1-0.6)*0.8*0.6+0.6*(1-0.8)*0.6+0.6*0.8*(1-0.6)=0.456

3. (1-0.8)*0.8*0.9+0.8*(1-0.8)*0.9+0.8*0.8*(1-0.9) =0.352

4. (1-0.3)*0.9*0.7+0.3*(1-0.9)*0.7+0.3*0.9*(1-0.7)=0.543

5. (1-0.1)*0.8*0.6+0.1*(1-0.8)*0.6+0.1*0.8*(1-0.6)=0.476

6.(1-0.5)*0.9*0.8+0.5*(1-0.9)*0.8+0.5*0.9*(1-0.8)=0.49

7.(1-0.7)*0.4*0.1+0.7*(1-0.4)*0.1+0.7*0.4*(1-0.1) =0.306

8. (1-0.7)*0.6*0.8+0.7*(1-0.6)*0.8+0.7*0.6*(1-0.8)=0.452

9.(1-0.1)*0.4*0.8+0.1*(1-0.4)*0.8+0.1*0.4*(1-0.8)=0.344

10. (1-0.8)*0.3*0.3+0.8*(1-0.3)*0.3+0.8*0.3*(1-0.3)=0.354

Estimated probability of snowing exactly two days=0.365+0.56+0.352+0.543+0.476+0.49+0.306+0.452+0.344+0.354)/10=0.4242

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