The only contents of a container are 10 disks that are each numbered with a different positive integ

migongoniwt

migongoniwt

Answered question

2022-06-18

The only contents of a container are 10 disks that are each numbered with a different positive integer from 1 through 10, inclusive. If 4 disks are to be selected one after the other, with each disk selected at random and without replacement, what is the probability that the range of the numbers on the disks selected is 7?

Answer & Explanation

Reagan Madden

Reagan Madden

Beginner2022-06-19Added 15 answers

You've got all the pieces, but there's a simpler way to put them together. There are, as you said, 3 possible ranges. There are ( 6 2 ) ways to choose 2 numbers in the range other than the endpoints, so there are 3 ( 6 2 ) possible choices. There are ( 10 4 ) equiprobable ways to choose 4 numbers, so the probability is
3 ( 6 2 ) ( 10 4 ) = 3 14
I'm having difficulty pinpointing your error. As I said, the method is confusing. I think the problem arises when you compute the probability of choosing 2 numbers from the middle of the range. You say the probability is 6 5 8 7 which is true, but this incldues both the possibility of choosing 5 and then 6 and the possibility of choosing 6 and then 5. When you then multiply by 4!, I think you're doubling up. I'm not 100% certain though.
Santino Bautista

Santino Bautista

Beginner2022-06-20Added 5 answers

Re the glitch in your answer, (3) x (1/10)(1/9)(6/8)(5/7) x (4!)
(6/8)(5/7) implies that the two discs within the limiters are already permuted, so the (4!) needs to be replaced by (4!/2!)
If you had written (2/10)(1/9) for the limiters, the final factor would become (4!)/(2!2!)

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?