A jar contains 4 red balls and 6 green balls Suppose you ran

A jar contains 4 red balls and 6 green balls. Suppose you randomly select 2 balls with replacement. Let X be the number of redd balls you select. Is X a binomial random variable? Yes. No. What is the probability that X=1, e.g. exactly one red ball is selected? A 0.16. B 0.48. C 0.36. D 0.4. What are EX) and VarX)? A 0.8,0.8 B 1.2,0.8 C 0.8,0.48 D 1.2,0.48. Now suppose these 2 balls are selected without replacement, then what is the probability that exactly one red ball ( and exactly one green ball) is selected? Please up to four decimal places. A 0.2667 B 0.3333 C 0.5333 D 0.4800

Solution

the jar contains 4 red balls and 6 green balls.

so the probability of obtaining a green ball=6/(4+6)=6/10=0.6

and the probability of obtaining a red ball=4/(4+6)=4/10=0.4

2 balls are drawn with replacement. X denote the number of red balls. so X can take the values 0,1,2.

15. yes X is a binomial random variable. as the balls are drawn with replacement each drawing are independent of each other. considering drawing a red ball as success, X denotes the number of successes in 2 independent trials. hence X follows a binomial distribution with parameters n=2,p=probability of drawing a red ball=0.4

hence pmf of X is P[X=x]=²C_x(0.4)^x(0.6)^2-x       x=0,1,2

16. so P[X=1]=²C₁(0.4)*(0.6)=0.48 [answer] [obtained by putting x=1 is the pmf]

17. for a binomial distribution E[X]=np=2*0.4=0.8

and variance is V[X]=np(1-p)=2*0.4*0.6=0.48

hence the correct option is option C [answer]

18. if these balls are drawn without replacement then X would not be a binomial distribution.

now from 10 balls ,2 balls can be selected in ¹⁰C₂ ways.

now to have one green ball and one red ball ,

one red balls must be chosen from 4 red balls in ⁴C₁ ways and 1 green ball must be chosen from 6 balls in ⁶C₁ ways.

hence one green ball and one red ball can be selected in ⁴C₁*⁶C₁ ways.

hence the required probability is ⁴C₁*⁶C₁/¹⁰C₂=0.5333   [answer option C]