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?
5/243
1/12
10/81
1/6
16/81
OE
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5/243
1/12
10/81
1/6
16/81
OE
[Reveal] Spoiler:
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 as, with n = 6 and k = 2, 6C2 = (30 / 2) = 15
To find number of ways to group remaining 4 experiments with exactly 2 bs, 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 (901/729) = (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 as, with n = 6 and k = 2, 6C2 = (30 / 2) = 15
To find number of ways to group remaining 4 experiments with exactly 2 bs, 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 (901/729) = (90/729) =10/81