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 Urn A has 3 red and 2 white marbles, ...
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Urn A has 3 red and 2 white marbles, urn B has 2 red and 5 white marbles. An urn is selected at random ; a marble is drawn, its colour noted and put into the other urn , then marble is drawn from the second urn. Find the probability that 1)both marbles drawn are red 2)both marbles drawn are white 3)marbles drawn are of same colour 4)second marble drawn is red 5)marbles drawn are of different colours 6)second marble drawn is white
The probability of selecting any urn (P) = 1/2.
Let A be the event of chosing urn A
Let B be the event of chosing urn B
The probability of drawing a red marble from urn A (R/A) = 3/5.
The probability of drawing a white marble from urn A (W/A) = 2/5
The probability of drawing a red marble from urn B (R/B) = 2/7
The probability of drawing a white marble from urn B (W/B) = 5/7
a. both marbles are red = (probability of selecting urn A *(probability selecting red marble from urn A)*(probability of selecting red marble from urn B after the marble selected from urn A is also put in the urn B) +(probability of selecting urn B *(probability selecting red marble from urn B)*(probability of selecting red marble from urn A after the marble selected from urn A is also put in the urn A)
Probability = (1/2*3/5*3/8 + 1/2*2/7*4/6)
= 0.1125+0.095
= 0.21
b. both marbles are white = Applying the same logic, we have
Probability = (1/2*2/5*6/8)+(1/2*5/7*3/6) = 0.33
c. both marbles are drawn of same color, so either both marbles are red or both marbles are white. Hence, probability = 0.21+0.33 = 0.54
d. second marble drawn is red, so the first marble drawn from either of the 2 urns can be either red or white. Hence, probability = (1/2*3/5*3/8 + 1/2*2/7*4/6)+(1/2*2/5*2/8)+(1/2*5/7*3/6) = 0.21+0.228 = 0.44
e. marble drawn are of different color Â– So, if first marble drawn is white, then the second is red and so on.
Probability = (1/2*2/5*2/8)+(1/2*5/7*3/6) + (1/2*3/5*5/8)+(1/2*2/7*2/6) = 0.228 +0.235 = 0.46
f. second marble drawn is white Â– So, the first one drawn can be either red or white. Hence,