sal_xcool wrote: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

OA C

ok, im confused as hell
somebody explain this to me please. is that considered 700+ question, because i really do hope so.

P(a) = P(b) = P(c) = 1/3
Number of ways to select the 2 occurrences for a, b and c out of 6 times = 6C2*4C2*2C2 = 15*6*1 = 90
In any one particular arrangement (say : "aabbcc") the probability = (1/3)^2 * (1/3)^2 * (1/3)^2 = (1/3)^6

Hence, the required probability = 90 * (1/3)^6 = 10/81 C

Alternate Approach:

You must have solved the dice problems. The dice has 6 faces. Think of a triangular solid shape here in this case (which is unbiased , as we know the probability of getting any side is equal = 1/3).

Total number of possible outcomes in 6 draws = 3^6
Desired number of outcomes = Number of ways of getting 2 a's, 2 b's and 2c's = 6C2*4C2*2C2 = 90

So, the required probability = Desired number of outcomes/Total number of possible outcomes = 90/3^6 = 10/81

sal_xcool wrote: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

OA C

ok, im confused as hell
somebody explain this to me please. is that considered 700+ question, because i really do hope so.

You can also solve using a variation of the permutations formula.

Probability = (# of desired outcomes)/(total # of possibilities)

Here, we're experimenting 6 times, and each time had 3 equally probable outcomes. So, the total number of possibilities is 3^6.

The number of ways to arrange n objects, when some of them are identical, is:

n!/r!s!t!...

in which n is the total number of objects and r, s and t are the number of duplicates.

We want to get two As, two Bs and two Cs, so our 6 objects to arrange are AABBCC.

Accordingly, there are:

6!/2!2!2! = 6*5*4*3*2/2*2*2 = 6*5*3 = 90 possible arrangements.

So, the probability of getting two of each is:

90/3^6 = 3^2*10/3^6 = 10/3^4 = 10/81

Please note that Stuart Kovinsky's approach and my alternate approach mentioned above are same (The only difference being the way to calculate the desired number of outcomes). Mentioning here for learning purposes.

sal_xcool wrote: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

OA C

ok, im confused as hell
somebody explain this to me please. is that considered 700+ question, because i really do hope so.

Probability-wise, a = 1/3, b = 1/3, and hence c = 1/3, and we have to find the probability of aabbcc occurring in any order, when the random experiment is conducted 6 independent times. There are 6! / (2! 2! 2!) = 90 different ways of getting aabbcc and the overall probability

= 90 × (1/3) ^6

= [spoiler]10/81


C[/spoiler]