0
$\begingroup$

I'm new to stats so I'm struggling to grasp the intuition behind permutations + probability. Would anybody be able to help me with parts (b) and (c) of this question?

Any help is greatly appreciated :)

A game requires you to match 10 words to 10 images (where each word correctly labels only one image). A machine randomly matches each word to a different image, where all possible labellings are equally likely.

What is the probability that the machine matches:

  • (a) all 10 words correctly?
  • (b) the first 7 words correctly?
  • (c) exactly 9 of the 10 words correctly?

This question is from a stats class problem set.

My work so far:

  • (a) $\frac{1}{10!}$ as $10!$ permutations are possible with only 1 having all correctly matched.
  • (b) I think $\frac{10!}{3!}$ but not 100% sure.
  • (c) Not too sure where to begin but I realise that if the last word is incorrectly matched then one of the 9 must also be incorrectly matched. What is the question instead said exactly 8 correct?
Improve this question
$\endgroup$
0

1 Answer 1

Reset to default
2
$\begingroup$

Hints:

  • How many equally likely permutations are there?

  • How many permutations get all $10$ words correct? What proportion of the total is this?

  • How many permutations get the first $7$ words correct and the last $3$ in any order? What proportion of the total is this?

  • If exactly $9$ out of $10$ words are correct then $1$ word is matched to a wrong image. What might have happened to the other word that should have been matched to that image?

Improve this answer
$\endgroup$
4
  • $\begingroup$ Thanks! Here's what I've got so far. • $10!$ permutations possible • $\frac{1}{10!}$ permutations gets all 10 correct • $\frac{10!}{7!}$ get the first 7 correct? • Not too sure for the last one $\endgroup$
    – lkdlst
    Commented Apr 22, 2021 at 10:16
  • $\begingroup$ $\frac{1}{10!}$ is correct. $\frac{10!}{7!}$ is not: if the first seven are correct, then you are counting the number of permutations of the last three. The final question has two possible answers: either the other word was matched correctly or it was matched incorrectly, and each of those has logical implications $\endgroup$
    – Henry
    Commented Apr 22, 2021 at 10:24
  • $\begingroup$ Right so $\frac{10!}{7!}$ should be $\frac{10!}{3!}$ and for the last one if it was matched incorrectly then that means it isn't possible to have the other 9 be correct - what if the question was instead asking about matching exactly 8 out of 10 correctly? $\endgroup$
    – lkdlst
    Commented Apr 22, 2021 at 10:28
  • $\begingroup$ You might want to read about rencontres numbers $\endgroup$
    – Henry
    Commented Apr 22, 2021 at 10:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.