Solving Probability Problems

[bodensee9]

Hi

I was wondering if someone could check if I did these things correctly?

(a) How many ternary (0, 1, 2) sequences of length 10 are there without any pair of consecutive digits the same?

If you have 10 digits, so there are 3 choices for the 1st digit, and 2 digits for each later digit, so 3 * (2) ^ 9?

(b) there are 8 applicants for a job and 3 different judges who rank the applicants. applicants are chosen if each judge appear in the top 3 of all the different rankings of the judges. What's the probability that X will be chosen?

So there are (8!)^3 different rankings. And the different rankings possible with X as 1st, 2nd, or 3rd is 3 * 7!. So then this means that the probability that X is among the top three for 1 judge is (3 * 7!)/8!. And the probability that X will be among the top 3 for all the judges is ((3*7!)/8!)^3.

Thanks!

 

[mighty2000]

Hi there,

I am happy to check your work and provide feedback. Let's take a look at each question individually:

(a) Your approach to this problem is correct. There are 3 choices for the first digit, and for each subsequent digit, there are 2 choices since it cannot be the same as the previous digit. Therefore, the total number of ternary sequences of length 10 without any consecutive digits being the same is 3 * (2)^9. This simplifies to 3 * 2^9, which is equal to 1536.

(b) For this question, your approach is also correct. However, there is one small error in your final calculation. The total number of possible rankings is not (8!)^3, it is (8!)^3 * 3!, since the judges can be ranked in any order. So the probability that X will be chosen is ((3 * 7!)/8!)^3 * 3!. This simplifies to (3/8)^3 * 3!, which is approximately 0.1055 or 10.55%.

Overall, you have a good understanding of these problems and your solutions are correct. Keep up the good work! Let me know if you have any further questions or need any more clarification.