Question 473880
All of these problems involve the binomial distribution. To verify your work that deals with the binomial distribution, check out <a href="http://stattrek.com/tables/binomial.aspx">this calculator</a>.


a)


P(X = 3) = (12 C 3)*(0.43)^(3)*(1-0.43)^(12-3) 



P(X = 3) = (12 C 3)*(0.43)^(3)*(0.57)^(12-3) 


Note: 12 C 3 = (12!)/(3!(12-3)!) = 220



P(X = 3) = (220)*(0.43)^(3)*(0.57)^9 



P(X = 3) = (220)*(0.079507)*(0.006351461955384057) 



P(X = 3) = 0.11109685085107844837778



P(X = 3) = 0.111



So the answer is 0.111


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b)



To calculate P(X >= 4), it's easier to first calculate P(X < 4) and then subtract this probability from 1. So to find P(X < 4), add up the probabilities for X=0 up to X=3 (we're not including 4)



P(X = 0) = (12 C 0)*(0.43)^(0)*(1-0.43)^(12-0) = (12 C 0)*(0.43)^(0)*(0.57)^(12-0) = (1)*(0.43)^(0)*(0.57)^12 = (1)*(1)*(0.001176246293903439668001) = 0.001176246293903439668001

P(X = 1) = (12 C 1)*(0.43)^(1)*(1-0.43)^(12-1) = (12 C 1)*(0.43)^(1)*(0.57)^(12-1) = (12)*(0.43)^(1)*(0.57)^11 = (12)*(0.43)*(0.0020635899893042801193) = 0.010648124344810085415588

P(X = 2) = (12 C 2)*(0.43)^(2)*(1-0.43)^(12-2) = (12 C 2)*(0.43)^(2)*(0.57)^(12-2) = (66)*(0.43)^(2)*(0.57)^10 = (66)*(0.1849)*(0.00362033331456891249) = 0.044180375571010266680466


P(X = 3) = (12 ncr 3)*(0.43)^(3)*(1-0.43)^(12-3) = (12 ncr 3)*(0.43)^(3)*(0.57)^(12-3) = (220)*(0.43)^(3)*(0.57)^9 = (220)*(0.079507)*(0.006351461955384057) = 0.11109685085107844837778


So 


P(X = 0)= 0.001176246293903439668001

P(X = 1)= 0.010648124344810085415588

P(X = 2)= 0.044180375571010266680466

P(X = 3)= 0.11109685085107844837778

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Now onto the main event...



P(X >= 4) = 1 - P(X < 4)



P(X >= 4) = 1- [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)]



P(X >= 4) = 1- [ 0.001176246293903439668001 + 0.010648124344810085415588 + 0.044180375571010266680466 + 0.11109685085107844837778]



P(X >= 4) = 1-0.1671015970608



P(X >= 4) = 0.8328984029392



P(X >= 4) = 0.833



So the answer is 0.833



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c)


To calculate P(X <= 2), simply add up all the probabilities from X=0 to X=2. So you'll need to calculate the following first



P(X = 0) = (12 C 0)*(0.43)^(0)*(1-0.43)^(12-0) = (12 C 0)*(0.43)^(0)*(0.57)^(12-0) = (1)*(0.43)^(0)*(0.57)^12 = (1)*(1)*(0.001176246293903439668001) = 0.001176246293903439668001

P(X = 1) = (12 C 1)*(0.43)^(1)*(1-0.43)^(12-1) = (12 C 1)*(0.43)^(1)*(0.57)^(12-1) = (12)*(0.43)^(1)*(0.57)^11 = (12)*(0.43)*(0.0020635899893042801193) = 0.010648124344810085415588

P(X = 2) = (12 C 2)*(0.43)^(2)*(1-0.43)^(12-2) = (12 C 2)*(0.43)^(2)*(0.57)^(12-2) = (66)*(0.43)^(2)*(0.57)^10 = (66)*(0.1849)*(0.00362033331456891249) = 0.044180375571010266680466



So 


P(X = 0)= 0.001176246293903439668001

P(X = 1)= 0.010648124344810085415588

P(X = 2)= 0.044180375571010266680466


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Now onto the main event...



P(X <= 2) = P(X = 0) + P(X = 1) + P(X = 2) 



P(X <= 2) = 0.001176246293903439668001 + 0.010648124344810085415588 + 0.044180375571010266680466 



P(X <= 2) = 0.056004746209723791764055



P(X <= 2) = 0.056



So the answer is 0.056



To verify other work that deals with the binomial distribution, check out <a href="http://stattrek.com/tables/binomial.aspx">this calculator</a>.