Answer to Question #328541 in Statistics and Probability for Sevi

Question

1. A fair die is rolled eight times. Find the probability that no more than 2

sixes come up.

2. A consulting group believes that 70% of the people in a certain city are

satisfied with their health coverage. Assuming that this is true, find the probability

that from a random sample of 15 people.

a. Exactly 10 are satisfied with their health coverage?

b. At least 2 are satisfied with their health coverage?

c. What is the expected number of people out of 15 that are satisfied with their

health coverage?

3. You observe that the number of telephone calls that arrive each day on

your mobile phone over a period of a year, and note that the average is 3 per day.

Let X be the number of calls that arrive in any one day. Then what is the probability

of receiving at most 2 calls in any one day? at least 2 calls in any one day?

Answer 1

$1)\space P(x=k)=C_n^kp^k(1-p)^{n-k}\\ \space P(x<2)=P(x=0)+P(x=1)=\\ =(\frac{5}{6})^8+8\cdot(\frac{1}{6})\cdot(\frac{5}{6})^7=\\ =0.604676902\approx60\%$

$2a)\space P(x=10)=C_{15}^{10}\cdot0.7^{10}\cdot0.3^5=\\ =\frac{15\cdot14\cdot13\cdot12\cdot11\cdot10\cdot9\cdot8\cdot7\cdot6}{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10}\cdot0.7^{10}\cdot0.3^5=\\ =\frac{15\cdot14\cdot13\cdot12\cdot11}{1\cdot2\cdot3\cdot4\cdot5}\cdot0.7^{10}\cdot0.3^5=\\ =0.206130381\approx21\%\\ 2b)\space P(x\ge2)=1-P(x=0)-P(x=1)=\\ =1-0.3^{15}-15\cdot0.7\cdot0.3^{14}=\\ =0.999999483\approx99.9999\%\\ 2c)\space E=pn=0.7\cdot15=10.5$

$3)\space P(x=k)=\frac{3^ke^{-3}}{k!}\\ P(x\le2)=P(x=0)+\\ +P(x=1)+P(x=2)=\\ e^{-3}(1+3+3^2/2)=\\ =0.423190081\approx42\%\\ P(x\ge2)=1-P(x\lt2)=\\ =1-(1+3)/e^3=0.800851727\approx80\%$