Probability of Rolling a Balanced Die 6 Times

[kingwinner]

1) A balanced die is tosssed six times, and the number on the uppermost face is recorded each time. What is the probability that the numbers recorded are 1, 2, 3, 4, 5, and 6 in any order?

My attempted answer is (6!)/(6^6), is it correct?


2) Suppose a die has been altered so that the faces are 1, 2, 3, 4, 5, and 5. If the die is tossed five times, what is the probability that the numbers recorded are 1, 2, 3, 4, and 5 in any oder?

No clue...can anyone help me, please?


3) We need to arrange 5 math books, 4 physics books and 2 statistics books on a shelf. How many possible arrangements exists so that books of the same subjects will lie side by side?

Is it 3! x (5! x 4! x 2!)?


Thanks for helping!

 

[Melawrghk]

Okay, let's see.
1) I don't think your attempted answer is correct. Think about it this way. You roll your die and want to get it to be "1", the chances of that happening are 1/6 (or 1!/6), same happens for all other numbers (2, 3, 4, 5 & 6). The total number of out comes (you got it right) is 6^6, because during each toss there are 6 possible outcomes. So, I think the answer should be (1!^6)/(6^6), or 1/(6^6). Try looking at the big picture, having your answer would mean that you would be able to get that sequence 240 times...

2) Use the same approach as before.
To get "1" in your first toss, the probability will be 1/6, same with 2, 3, 4. But when we get to 5, the picture changes. Instead of probability being 1/6, it is now 2/6, because there are two faces labeled as "5". So the probability of that sequence occurring is
1/6 * 1/6 * 1/6 * 1/6 * 2/6 * 2/6
^#1 ^#2 ^#3 ^#4 ^#5 ^second #5

3) Yes, I think that's right.

 

[Dick]

Okay, let's see.
1) I don't think your attempted answer is correct. Think about it this way. You roll your die and want to get it to be "1", the chances of that happening are 1/6 (or 1!/6), same happens for all other numbers (2, 3, 4, 5 & 6). The total number of out comes (you got it right) is 6^6, because during each toss there are 6 possible outcomes. So, I think the answer should be (1!^6)/(6^6), or 1/(6^6). Try looking at the big picture, having your answer would mean that you would be able to get that sequence 240 times...

2) Use the same approach as before.
To get "1" in your first toss, the probability will be 1/6, same with 2, 3, 4. But when we get to 5, the picture changes. Instead of probability being 1/6, it is now 2/6, because there are two faces labeled as "5". So the probability of that sequence occurring is
1/6 * 1/6 * 1/6 * 1/6 * 2/6 * 2/6
^#1 ^#2 ^#3 ^#4 ^#5 ^second #5

3) Yes, I think that's right.

kingwinner's solution is completely correct for 1). The number of permutations of the six numbers is 6!. The probability of each permutation occurring is 1/6^6. Answer: 6!/6^6. Think of the second problem the same way. How many permutations, what's the probability of each permutation? No arguments with the last one.