Probability Question

n=5, p=0.4, q=0.6

1) possible values are, X=1, 2, 3 or 4. i am not sure about this question but i think there are 4 defective prints therefore X can be 1, 2, 3 or 4 when choosing at random for 5

2) p(X=2) = 0.3456

3) p(X<2) = 0.33696

Ps, i am just using a graphic calculator to get the answers for Q2 and Q3

2010-11-26 05:32:48 補充：
In Question 1, should be X = 0, 1, 2, 3,4
In Question 3,
p(X<2) not only = p(X=1)
p(X<2) should be equal to p(X=1) + p(X=0)

2010-11-26 05:35:45 補充：
Therefore, I think Eric did not really get the right answers for you

Probability of defective printer = 4 /10 = 0.4
Probability of not defective printer = 6 /10 = 0.61. what are the possible values of XNot sure what question asked?
If all 5 printers are in good working condition, possbile value of X = 0.2. what is the probability that the compnay would get 2 defective printers among the 5 chosen?
nCr=(2C5)=10Ans: 0.4 x 0.4 x 0.6 x 0.6 x 0.6 x 10
= 0.3456
3. what is the probability that the compnay would get fewer than 2 defective printers among the 5 chosen?nCr=(1C5)=5Ans: 0.4 x 0.6 x 0.6 x 0.6 x 0.6 x 5
= 0.2592

2010-11-24 18:35:43 補充：
1. what are the possible values of X
X = is the random variable representing the number of defective printers that the company has chosen,

possbile value of X = 0, for all 5 printers are in good working condition.