Question 730800
Both parts will use the binomial distribution probability formula, which is


P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)


for both parts A and B, n = 10 and p = 0.8


A)


For this part A) only, k = 8


P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)


P(X = 8) = (10 C 8)*(0.8)^(8)*(1-0.8)^(10-8)


P(X = 8) = (10 C 8)*(0.8)^(8)*(0.2)^(10-8)


P(X = 8) = (45)*(0.8)^(8)*(0.2)^2


P(X = 8) = (45)*(0.167772)*(0.04)


P(X = 8) = <font color="red">0.30199</font>


So the probability that exactly 8 students sent and received text messages is roughly <font color="red">0.30199</font>


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B)


For this part, we need to find P(X = 7), P(X = 8), P(X = 9), and P(X = 10). We will add them up to calculate P(X >= 7)



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P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)


P(X = 7) = (10 C 7)*(0.8)^(7)*(1-0.8)^(10-7)


P(X = 7) = (10 C 7)*(0.8)^(7)*(0.2)^(10-7)


P(X = 7) = (120)*(0.8)^(7)*(0.2)^3


P(X = 7) = (120)*(0.209715)*(0.008)


P(X = 7) = 0.201327


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P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)


P(X = 8) = (10 C 8)*(0.8)^(8)*(1-0.8)^(10-8)


P(X = 8) = (10 C 8)*(0.8)^(8)*(0.2)^(10-8)


P(X = 8) = (45)*(0.8)^(8)*(0.2)^2


P(X = 8) = (45)*(0.167772)*(0.04)


P(X = 8) = 0.30199


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P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)


P(X = 9) = (10 C 9)*(0.8)^(9)*(1-0.8)^(10-9)


P(X = 9) = (10 C 9)*(0.8)^(9)*(0.2)^(10-9)


P(X = 9) = (10)*(0.8)^(9)*(0.2)^1


P(X = 9) = (10)*(0.134218)*(0.2)


P(X = 9) = 0.268435


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P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)


P(X = 10) = (10 C 10)*(0.8)^(10)*(1-0.8)^(10-10)


P(X = 10) = (10 C 10)*(0.8)^(10)*(0.2)^(10-10)


P(X = 10) = (1)*(0.8)^(10)*(0.2)^0


P(X = 10) = (1)*(0.107374)*(1)


P(X = 10) = 0.107374


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So in short, we found this


P(X = 7)= 0.201327
P(X = 8)= 0.30199
P(X = 9)= 0.268435
P(X = 10)= 0.107374


They add to...


P(X >= 7) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)


P(X >= 7) = 0.201327 + 0.30199 + 0.268435 + 0.107374 


P(X >= 7) = <font color="red">0.879126</font>


So the probability that at least 7 of these students send and receive text messages is approximately <font color="red">0.879126</font>