3. An urn contains two red balls, three green balls and one white ball. Three ba

Question

3. An urn contains two red balls, three green balls and one white ball. Three balls are drawn from the urn without replacement, in sequence, and only their colors noted. (a) List the sample space. (Use r red, g 'green white.) (b) List the outcomes in each of these events: A "first bal is white", B "exactly two red balls are drawn" and C"the three balls are different colors" (c) Find AnB, BUC and Anc (d) Use the multiplication rule for counting to find P((g.r.r)) and P((.r.9)). Are the outcomes in this sample space equally likely? Explain (e) Find P(A), P(B), P(AnB) and P(AUB

Explanation / Answer

a)sample space S={ ggg ; ggr ;ggw ; grg ;gwg ;rgg ;wgg ;rrg ;rrw ; rgr ;rwr ; grr ;wrr ;rgw;rwg ; grw ;gwr;

wrg;wgr}

b)

sample space for A ={ wgg ;wrr ;wgr ;wrg]

sample space for B ={ rrg ;rrw ; rgr ;rwr ; grr ;wrr }

sample space for C ={ rgw;rwg ; grw ;gwr; wrg;wgr}

c)

A n B = { wrr }

B u C ={ rrg ;rrw ; rgr ;rwr ; grr ;wrr ; rgw;rwg ; grw ;gwr; wrg;wgr}

A n Cc ={ wgg ;wrr }

d)

P({ g,r,r}) =(3/6)*(2/5)*(1/4)=1/20

P({g,r,g}) =(3/6)*(2/5)*(2/4)=1/10

the events are not equally liekly as for there are 3 green balls and 2 red balls hence probability of thee events are different

e)

P(A)=(1/6) (as from 6 balls 1 is white)

P(B)=P(first two are red and third is green or white+first and third are red and second is green or white+second and third are red and first is green or white)

=(2/6)*(1/5)*(4/4)+(2/6)*(4/5)*(1/4)+(4/6)*(2/5)*(1/4)=1/5

P(A n B)=P(first is white and rest 2 are red)=(1/6)*(2/5)*(1/4)=1/60

P(A u B)=P(A)+P(B)-P(A n B)=(1/6)+(1/5)-1/60 =(10+12-1)/60=21/60

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