Help with Binomial Distribution please!?

The question: Suppose that in a certain area 9 out of 10 households have a VCR. Let x denote the number among four randomly selected households that have a VCR, so x is a binomial random variable with n = 4 and p = .9.





a. calculate p(2) and interpret


b. calculate p(4)


c. determine P(x equal or less than 3)





I would really like to understand how this works! I have the equation but I am not sure how to factor it. Please be detailed in your response. thanx

Help with Binomial Distribution please!?
♠ if p=0.9 then q=1-p=0.1 means probability that a man taken at random has no VCR, while probability p=0.9 means he has a VCR. So probability either he has or he has not = p+q =1;


a)♣ now I take 2 men at random then probability of all possible combinations has/hasn’t for 2 men is also =1; and 1=1^2 =(p+q)^2 = p^2 +2pq +q^2, where


p^2 means probability of Ken %26amp; Dan each having a VCR;


2pq means probability EITHER Ken has %26amp; Dan hasn’t OR Dan has %26amp; Ken not;


q^2 means probability of Ken %26amp; Dan each having no VCR;


thus your p(2) = p^2 =0.9^2 =0.81 probability; for 2 men;


b)♦ thus your p(4) = p^4 =0.9^4 =0.6561 probability; for 4 men;


c)♥ we take 3 men at random then 1^3 =(p+q)^3 =


= p^3 +3p^2q +3pq^2 +q^3;


p(x%26lt;=3) is probability of one or more VCR for 3 men


and q^3 means probability that 3 men have no VCR;


thus p(x%26lt;=3) = 1-q^3 = 0.999 = p^3 +3p^2q +3pq^2; check it!

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