probability

Question

There are $N$ persons sitting in a row. Two of them are selected at random.The probability that two selected persons are not together ?

Answer 1

Now n-persons are sitting in a row so we can select two people who are sitting consecutively in n-1 ways. Total number of ways of selecting two people is ⁿC_2.

So the required probability is

( ⁿC₂ - (n-1) ) / ⁿC₂

Comments

if he chooses corner seats then 2*(n-2)/n *(n-1) [because except the chosen and its adjacent one.others can be chosen]

if he chooses except corners then (n-2)*(n-3)/n * (n-1) [because except the chosen and two of its adjacent persns,others can be chosen]

so, 2*(n-2)/n * (n-1) + (n-2)*(n-3)/n * (n-1)

can we do like this??

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@Akriti, In your approach lets say a,b,c,d are seated in a row then by case one where corner seat is chosen pair can be a,c and in case two where non-corner seated is chosen lets say we chose c and then we choose a so we get c,a , so I think single possibility is counted twice.

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i got your point..but i was getting the same answer when i put n=6 in your and my method.