+0  
 
+1
45
2
avatar+184 

Let S be the set of permutations of \((1,2,3,4,5,6)\) whose first term is not 1.

If we choose a permutation at random from S, what is the probability that the third term is equal to 3?

________________________________________

I am quite confused on how to start, so how do i start? If possible, can you please provide multiple hints, or the solution?

Max0815  Sep 16, 2018
 #1
avatar+2323 
+3

let's first figure out how many permutations do have the 3rd term equal to 3

 

The first number we have 4 choices (not 1 or 3), the second number we have 4 choices, the 3rd number is 3, the 4th there are 3 choices, 5th 2 choices.

 

This is 4 x 4! = 96

 

The total number of permutations in S is just 5 x 5! = 600 (do you see why?)

and thus

 

P[term 3 = 3] = 96/600 = 4/25

Rom  Sep 16, 2018
 #2
avatar+184 
+1

Thank you very much :)

Max0815  Sep 16, 2018

26 Online Users

avatar
avatar
avatar
avatar
avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.