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Updated on: 12 Jan 2014, 18:34
Originally posted by goodyear2013 on 12 Jan 2014, 16:44.
Last edited by Bunuel on 12 Jan 2014, 18:34, edited 1 time in total.
Last edited by Bunuel on 12 Jan 2014, 18:34, edited 1 time in total.
Edited the question.
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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In a certain random experiment, exactly one of the outcomes A, B, and C will occur. In each random experiment, the probability that outcome A will occur is 1/3 , and the probability that outcome B will occur is 1/3. What is the probability that when the random experiment is conducted 6 independent times, each of outcomes A, B, and C will occur twice?
A. 5/243
B. 1/12
C. 10/81
D. 1/6
E. 16/81
OE
A. 5/243
B. 1/12
C. 10/81
D. 1/6
E. 16/81
OE
In each random experiment, sum of probabilities that outcomes a, b, and c occur is 1.
Probability that each of c occurs is 1/3 since a and b is each 1/3.
Probability that any specified sequence of outcomes in 6 independent repetitions of random experiment occurs is (1/3)^6= (1/729)
Probability that each of events a, b, and c occurs twice is equal to number of different possible sequences of 6 outcomes where each of outcomes a, b, and c occurs twice, multiplied by 1/729
Number of different possible sequences of 6 outcomes in 6 independent random experiments, where each of outcomes a, b, and c occurs exactly twice, is number of different ways to group 6 experiments such that 2 have outcome a, 2 have outcome b, and 2 have outcome c.
To find number of groupings with exactly 2 a’s, with n = 6 and k = 2, 6C2 = (30 / 2) = 15
To find number of ways to group remaining 4 experiments with exactly 2 b’s, with n = 4 and k = 2,
4C2 = 6
Since assigned exactly 2 experiments with outcome a and exactly 2 experiments with outcome b, remaining 2 experiments must be outcome c, so no further application of formula is necessary.
Thus, total number of desired outcomes is product of 2 combinations calculations, 15 x 6 = 90.
Probability that each of outcomes a, b, and c will occur twice is (90×1/729) = (90/729) =10/81
Probability that each of c occurs is 1/3 since a and b is each 1/3.
Probability that any specified sequence of outcomes in 6 independent repetitions of random experiment occurs is (1/3)^6= (1/729)
Probability that each of events a, b, and c occurs twice is equal to number of different possible sequences of 6 outcomes where each of outcomes a, b, and c occurs twice, multiplied by 1/729
Number of different possible sequences of 6 outcomes in 6 independent random experiments, where each of outcomes a, b, and c occurs exactly twice, is number of different ways to group 6 experiments such that 2 have outcome a, 2 have outcome b, and 2 have outcome c.
To find number of groupings with exactly 2 a’s, with n = 6 and k = 2, 6C2 = (30 / 2) = 15
To find number of ways to group remaining 4 experiments with exactly 2 b’s, with n = 4 and k = 2,
4C2 = 6
Since assigned exactly 2 experiments with outcome a and exactly 2 experiments with outcome b, remaining 2 experiments must be outcome c, so no further application of formula is necessary.
Thus, total number of desired outcomes is product of 2 combinations calculations, 15 x 6 = 90.
Probability that each of outcomes a, b, and c will occur twice is (90×1/729) = (90/729) =10/81
12 Jan 2014, 18:37
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goodyear2013
Given that P(A) = P(B) = P(C) = 1/3.
P(AABBCC) = \((\frac{1}{3})^6*\frac{6!}{2!2!2!} = \frac{10}{81}\), we multiply by 6!/(2!2!2!) because AABBCC scenario can occur in several ways: AABBCC, AABCBC, AACBBC, ... (the number of permutations of 6 letters AABBCC is 6!(2!2!2!)).
Answer: C.
General Discussion
29 Oct 2014, 22:06
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Number of ways in which a group of 2A,2B,2C can be formed in any order = 6!/(2!*2!*2!) [ formula for total of 6 items with 2 each of similar kind)
Total number of ways in which the die can be thrown independently = _ _ _ _ _ _ ; there are 6 places .First can be filled in 3 different ways A/B/C; Second in 3 diff ways again and so on.
So total number of ways = 3X3X3X3X3X3 = 3^6
There fore probability = 6!/(2!x2!x2!) by 3^6 = 90/3^6 = 10/81 Answer C
Hope this is clear !
Total number of ways in which the die can be thrown independently = _ _ _ _ _ _ ; there are 6 places .First can be filled in 3 different ways A/B/C; Second in 3 diff ways again and so on.
So total number of ways = 3X3X3X3X3X3 = 3^6
There fore probability = 6!/(2!x2!x2!) by 3^6 = 90/3^6 = 10/81 Answer C
Hope this is clear !
