Two dice are thrown simultaneously. Find the probability of getting a multiple of 2 one dice and a multiple of 3 on an other dice.
Answers
Answer: Elementary events associated to the random experiment of throwing two dice are:
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
∴ Total number of elementary events = 6 X 6 = 36
Let A be the event of getting a multiple of 2 on one die and a multiple of 3 on the other. Then, the elementary events favourable to A are:
(2, 3), (2, 6), (4,3), (4, 6), (6, 3), (6, 6), (3, 2), (3, 4), (3, 6), (6, 2), (6, 4).
∴ favourable number of elementary events = 11
Hence, required probability =
11/36
Answer
Probability is 1/6
Explanation
As 2 dice are thrown simultaneously. Therefore, the sample space will be
SS= {1 1, 1 2, 1 3 , 1 4, 1 5, 1 6, 2 1, 2 2, 2 3, 2 4, 2 5, 2 6, 3 1, 3 2, 3 3, 3 4, 3 5, 3 6, 4 1, 4 2, 4 3, 4 4, 4 5, 4 6, 5 1, 5 2, 5 3, 5 4, 5 5, 5 6, 6 1, 6 2, 6 3, 6 4, 6 5, 6 6}= 36= number of sample space
therefore, the events in which the first die has multiple of 2 and the second die has multiple of three are= 2 3, 2 6, 4 3, 4 6, 6 3, 6 6= 6
Hence the probability = events / number of sample space
6/36=1/6
probability = 1/6
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