Consider two boxes containing marbles of differing colors: BOXA 2 red BOXB 3 red
ID: 3302322 • Letter: C
Question
Consider two boxes containing marbles of differing colors: BOXA 2 red BOXB 3 red Total: 5 marbles Total: 4 marbles The exterior of the boxes are identical and we are never allowed to inspect the interior of the boxes. The boxes are randomized so that we don't know which box is which. We will make one draw at random from EACH box. Answer the following: Define events A: the first draw will be made from box A Bi: the first draw will be made from box B Ri: the first draw will be red Rz: the second draw will be red Gi: the first draw will be green G2: the second draw will be green Given the first ball drawn was RED, find the probability that the draw was made from box A a. P(4 R)- b. Given the first draw was RED, find the probability the second draw will be GREEN. (* keep in mind we draw once from each box*) P(G, R) | Find the probability that both marbles drawn will be the same color c. Note: this is not a conditional probabilityExplanation / Answer
a) P(R1) =P(first box is A nd red ball drawn +first box is B and red ball drawn )
=P(A)*P(R1|A)+P(B)*P(R1|B) =(1/2)*(2/5)+(1/2)*(3/4) =0.575
therefore P(A1|R1) =P(A)*P(R1|A)/P(R1) ==(1/2)*(2/5)/0.575 =0.347826
b)
P(G2|A1) =(P(G2nA1))/P(A1) =P(P(A1)*P(R1|A1)*P(G2|B)+P(B1)*P(R1|B1)*P(G2|A))/P(R1)
=((1/2)*(2/5)*(1/4)+(1/2)*(3/4)*(3/5))/0.575 =0.478261
c)probability that both marble drawn will be same color =P(A)*P(R1|A)*P(R2|B)+P(B)*P(R1|B)*P(R2|A)+P(A)*P(G1|A)*P(G2|B)+P(B)*P(G1|B)*P(G2|A)
=(1/2)*(2/5)*(3/4)+(1/2)*(3/4)*(2/5)+(1/2)*(3/5)*(1/4)+(1/2)*(1/4)*(3/5) =0.45