Problem Solving

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 6/81

OA C


Is there any simple method to approach this problem. Mine was a lengthy one.

jsasipriya 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 , and the probability that outcome b will occur is . 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 6/81

OA C


Is there any simple method to approach this problem. Mine was a lengthy one.

Can you please recheck the question??

jsasipriya 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 , and the probability that outcome b will occur is . 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 6/81

OA C


Is there any simple method to approach this problem. Mine was a lengthy one.

I think u missed out the probability of occurrence of a, b etc.. please repost the question...if you provide the data then its easy to solve....

I'm sorry. I missed a few data. It is now updated.

I got a (clearly) incorrect answer, but I can't find the flaw in my method either

P(a) = P(b) = P(c) = 1/3.

Now, lets say take any one of the desired outcomes (for instance aabbcc).

Now, probability of this happening is 1/(3^6). This means that probability of one of the favourable conditions is 1/(3^6).

Further, we can have 6! arrangements of aabbcc, and probability of each arrangement will be 1/(3^6), which means total probability of having 2 of each a,b and c should be 6!/(3^6) = 80/81

Clearly wrong, but how?

Can one of the experts please help with this one?

jsasipriya 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 6/81

OA C


Is there any simple method to approach this problem. Mine was a lengthy one.

jsasipriya 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 6/81

OA C


Is there any simple method to approach this problem. Mine was a lengthy one.

Probability of getting AABBCC = (1/3) ^ 6.
Number of ways to arrange AABBCC = 6!/(2!)(2!)(2!) = 90
(1/3)^6 * 90 = 10/81.

Quick and easy if you understand all the methodology that I used. If you need further explanation, please let me know.

GMATGuruNY wrote:

jsasipriya 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 6/81

OA C


Is there any simple method to approach this problem. Mine was a lengthy one.

Probability of getting AABBCC = (1/3) ^ 6.
Number of ways to arrange AABBCC = 6!/(2!)(2!)(2!) = 90
(1/3)^6 * 90 = 10/81.

Quick and easy if you understand all the methodology that I used. If you need further explanation, please let me know.

Ah, thanks. I forgot to divide by 2!2!2! . D'oh!

GMATGuruNY wrote:

jsasipriya 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 6/81

OA C


Is there any simple method to approach this problem. Mine was a lengthy one.

Probability of getting AABBCC = (1/3) ^ 6.
Number of ways to arrange AABBCC = 6!/(2!)(2!)(2!) = 90
(1/3)^6 * 90 = 10/81.

Quick and easy if you understand all the methodology that I used. If you need further explanation, please let me know.

GMATGuruNY,
Could you please explain on the number of ways?

Thanks

jsasipriya wrote:

GMATGuruNY wrote:

jsasipriya 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 6/81

OA C


Is there any simple method to approach this problem. Mine was a lengthy one.

Probability of getting AABBCC = (1/3) ^ 6.
Number of ways to arrange AABBCC = 6!/(2!)(2!)(2!) = 90
(1/3)^6 * 90 = 10/81.

Quick and easy if you understand all the methodology that I used. If you need further explanation, please let me know.

GMATGuruNY,
Could you please explain on the number of ways?

Thanks

A quick refresher on permutations:

How many ways can the letters ABC be arranged if no letters are repeated?

We have 3 choices for the 1st position.
We have 2 choices left for the 2nd position. (Because we used 1 of our 3 choices for the first position, leaving us only 2 choices for the 2nd position.)
We have 1 choice left for the last position. (Because we used 2 of our 3 choices for the first two positions, leaving us only 1 choice for the last position.)

Now we multiply the number of choices for each position:

3 * 2 * 1 = 3! = 6 possible arrangements.

So the total number of ways n elements can be arranged is n!

Now a harder question:

How many ways can the letters AAB be arranged if the letter A must appear exactly twice in the arrangement?

Normally, the number of arrangements would be 3!. But we have to account for the repeated A, which will reduce the number of unique arrangements. (If we wrote out all the possible unique arrangements, we'd see that there are only 3: AAB, ABA, and BAA.)

When an element is repeated, use the following formula:

(total number of possible arrangements)/(number of repetitions!).

In AAB we have 3 elements and 2 repetitions of the letter A, so we have to divide by 2!:

3!/2!= 3.

When more than one element is repeated, we have to account for each repeated element by dividing by (number of repetitions!) for each repeated element.

In the problem above, we need to determine how many ways we can arrange AABBCC.

We have 6 elements. We have to divide by 2! to account for the 2 repetitions of the letter A, by another 2! to account for the 2 repetitions of the letter B, and by another 2! to account for the 2 repetitions of the letter C:

6!/(2! * 2! * 2!) = 90.