SOLUTION: 65% of men consider themselves football fans,you randomly select 8 men and ask each if he is a football fan.Find the probability that 1.at least 5 are football fans 2.less than f

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Question 1086877: 65% of men consider themselves football fans,you randomly select 8 men and ask each if he is a football fan.Find the probability that
1.at least 5 are football fans
2.less than five are football fans.

Answer by jim_thompson5910(35256) (Show Source):
You can put this solution on YOUR website!
Let's compute the binomial probabilities for k = 5 through k = 8 (where k is an integer)
For each case, the probability of success is p = 0.65 and the sample size is n = 8
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For k = 5, we have this combination
n C k = (n!)/(k!*(n-k)!)
8 C 5 = (8!)/(5!*(8-5)!)
8 C 5 = (8!)/(5!*3!)
8 C 5 = (8*7*6*5!)/(5!*3!)
8 C 5 = (8*7*6)/(3!)
8 C 5 = (8*7*6)/(3*2*1)
8 C 5 = 336/6
8 C 5 = 56

Leading to this binomial probability
P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 5) = (8 C 5)*(0.65)^(5)*(1-0.65)^(8-5)
P(X = 5) = (8 C 5)*(0.65)^(5)*(0.35)^(3)
P(X = 5) = (56)*(0.65)^(5)*(0.35)^3
P(X = 5) = (56)*(0.1160290625)*(0.042875)
P(X = 5) = 0.2785857790625
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For k = 6, we have this combination
n C k = (n!)/(k!*(n-k)!)
8 C 6 = (8!)/(6!*(8-6)!)
8 C 6 = (8!)/(6!*2!)
8 C 6 = (8*7*6!)/(6!*2!)
8 C 6 = (8*7)/(2!)
8 C 6 = (8*7)/(2*1)
8 C 6 = 56/2
8 C 6 = 28

Leading to this binomial probability
P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 6) = (8 C 6)*(0.65)^(6)*(1-0.65)^(8-6)
P(X = 6) = (8 C 6)*(0.65)^(6)*(0.35)^(2)
P(X = 6) = (28)*(0.65)^(6)*(0.35)^2
P(X = 6) = (28)*(0.075418890625)*(0.1225)
P(X = 6) = 0.25868679484375
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For k = 7, we have this combination
n C k = (n!)/(k!*(n-k)!)
8 C 7 = (8!)/(7!*(8-7)!)
8 C 7 = (8!)/(7!*1!)
8 C 7 = (8*7!)/(7!*1!)
8 C 7 = (8)/(1!)
8 C 7 = (8)/(1)
8 C 7 = 8/1
8 C 7 = 8

Leading to this binomial probability
P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 7) = (8 C 7)*(0.65)^(7)*(1-0.65)^(8-7)
P(X = 7) = (8 C 7)*(0.65)^(7)*(0.35)^(1)
P(X = 7) = (8)*(0.65)^(7)*(0.35)^1
P(X = 7) = (8)*(0.04902227890625)*(0.35)
P(X = 7) = 0.1372623809375
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For k = 8, we have this combination
n C k = (n!)/(k!*(n-k)!)
8 C 8 = (8!)/(8!*(8-8)!)
8 C 8 = (8!)/(8!*0!)
8 C 8 = (1)/(1*0!)
8 C 8 = (1)/(1*1)
8 C 8 = (1)/(1)
8 C 8 = 1

Leading to this binomial probability
P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 8) = (8 C 8)*(0.65)^(8)*(1-0.65)^(8-8)
P(X = 8) = (8 C 8)*(0.65)^(8)*(0.35)^(0)
P(X = 8) = (1)*(0.65)^(8)*(0.35)^0
P(X = 8) = (1)*(0.0318644812890625)*(1)
P(X = 8) = 0.0318644812890625
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We do not need to compute the probabilities for k = 0 to k = 4, but for the sake of completeness the probabilities are shown in the table below

Full Binomial Distribution Table

X = k P(X = k)

0 0.0002251875390625

1 0.0033456434375

2 0.02174668234375

3 0.0807733915625

4 0.187509658984375

5 0.2785857790625

6 0.25868679484375

7 0.1372623809375

8 0.0318644812890625

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We'll use that table to answer problem 1. Add up the probabilities for k = 5, k = 6, k = 7, k = 8 to get

Answer for problem one: 0.70639943613281
Round that however you need to

The events "less than 5" and "at least 5" are complementary. One or the other must happen. Therefore their probabilities add to 1

Use the result from problem 1 to get

Answer for problem two: 0.29360056386719
Round that however you need to