Quadrilateral A’B’ C’ D ‘ is the result of reflecting quadrilateral ABCD across line l
1 answer:
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Drawing the tree diagram of the problem, we see that there are 6 branches for drawing from Urn A, and then 4 branches for each of the 6 branches, for drawing from Urn B.
This means that there are 6*4=24 possible outcomes, which could be listed as
{(1, 1), (1, 2), (1, 3), (1, 4), (2, 1)... (6, 3), (6, 4)},
where the first coordinates represent drawing from urn A, and the second coordinates, drawing from urn B.
(4, 2) is one of these 24, so the probability is 1/24
remark, this problem could also have been solved by the multiplication principle of the probabilities of separate events, that is 1/6 (probability of drawing 4 from Urn A) times 1/4 (probability of drawing 2 from B) = 1/24
Answer: 1/24
This means that there are 6*4=24 possible outcomes, which could be listed as
{(1, 1), (1, 2), (1, 3), (1, 4), (2, 1)... (6, 3), (6, 4)},
where the first coordinates represent drawing from urn A, and the second coordinates, drawing from urn B.
(4, 2) is one of these 24, so the probability is 1/24
remark, this problem could also have been solved by the multiplication principle of the probabilities of separate events, that is 1/6 (probability of drawing 4 from Urn A) times 1/4 (probability of drawing 2 from B) = 1/24
Answer: 1/24