P1.5 A desk drawer contains 1 black, 2 blue, and 3 red pens. We draw pens from t

Question

P1.5 A desk drawer contains 1 black, 2 blue, and 3 red pens. We draw pens from the drawer one by one without replacement until two pens with different colors are drawn. Let random variable X represent the number of pens drawn in this experiment. a. What is the range of X? (i.e. all the possible values X can take) b. Find and draw the probability mass function (PMF) of X, Px(x) - P(X -x) c. Determine the probability that we draw at least 3 pens, P(X23) d. What is possible values X could take given that the first pen we drew was black.

Explanation / Answer

a)

range of X ={x:x=2,3,4} (as for two different pen minimum number required is 2 as well cause there are maximum 3 pens of same color (red); hence if we pick first 3 red then 4th must be of different color)

b)

below is pmf of X:

P(X=2)=P(1st black and second blue)+P(1st black and second red)+P(1st blue and second black)+P(1st blue and second red)+P(1st red and second black)+P(1st red and second blue)

P(X=2)=(1/6)*(2/5)+(1/6)*(3/5)+(2/6)*(1/5)+(2/6)*(3/5)+(3/6)*(1/5)+(3/6)*(2/5)=11/15

P(X=3)=P(first 2 blue and third black)+P(first 2 blue and third red)+P(first 2 red and third black)+P(first 2 red and third blue)

P(X=3)=(2/6)*(1/5)*(1/4)+(2/6)*(1/5)*(3/4)+(3/6)*(2/5)*(1/4)+(3/6)*(2/5)*(2/4) =13/60

P(X=4)=P(first 3 red and 4th black)+P(first 3 red and 4th blue)

P(X=4)=(3/6)*(2/5)*(1/4)*(2/3)+(3/6)*(2/5)*(1/4)*(1/3) =1/20

c)

P(at least 3 pens)=P(X>=3) =P(X=3)+P(X=4) =(13/60)+(1/20)=16/60=4/15

d)

if first pen is black ; then second can be of red or blue as there is only 1 balck pen which has beeen taken

therefore X could take only 2 as a value.

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