Please provide as much detail as you can. Thanks in advance Three balls are to b

Question

Please provide as much detail as you can. Thanks in advance

Three balls are to be selected without replacement from an urn that contains 6 red, 5 green, and4 black balls. Let Yi and Y2 be random variables for the number of red and green balls selected, respectively. (a) Give the expression for p(yi,y2), the joint probability function for Y. and Y2 (b) Give the marginal probability functions for Y and Y2. (c) Find the probability that at least one red ball and exactly one green ball is selected (d) Find the probability that exactly one green ball is selected given that at least one red ball is selected. (e) Repeat (a) to (d), but assume that sampling is with replacement.

Explanation / Answer

Solution. a

For finding the joint probability of p(Y1,Y2) first we have to find the individual probabilty of Y1 and Y2

Y1 is is the probabilty of selecting red ball

P(Y1)=6/15*5/14*4/13

P(Y1)=4/91

Y2 is the probabilty of selecting green ball

P(Y2)=5/15*4/14*3/13

P(Y2)=2/91

P(Y1,Y2)=P(Y1)*P(Y2)

P(Y1,Y2)=4/91*2/91

P(Y1,Y2)=8/8281

b)

Marginal probability of Y1 is 4/91.

Marginal probabilty of Y2 is 2/91.

c)

Let Z be the probability of at least one red and one green

selected,

P(Z)=6/15*5/14*4/13

P(Z)=4/91

d)

Let A be the probability of at one green ball selected given that at least one red ball selected

P(A)=6/14*4/13

P(A)=12/9

e)

calculating probability with replacement

P(Y1)=6/15*6/15*6/15

P(Y1)=216/3375

P(Y2)=5/15*5/15*5/15

P(Y2)=1/27

P(Y1,Y2)=P(Y1)*P(Y2)

P(Y1,Y2)=216/3375*1/27

P(Y1,Y2)=8/3375

P(A)=6/15*4/15

P(A)=24/225

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