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What is the probability that there is an edge in an undirected random graph having 8 vertices?

  1. 1
  2.  1/8  
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2 Answers

5 votes
5 votes

Probability of taking $0$ vertex among $8$ vertices is $^{8}\textrm{C}_{0}$ [means no edge]

Probability of taking $1$ vertex among $8$ vertices is $^{8}\textrm{C}_{1}$ [means no edge]

Probability of taking $2$ vertex among $8$ vertices is $^{8}\textrm{C}_{2}$ [means $1$ edge]

.......

Now we can take any number of edges

So, number of vertices we can select=$^{8}\textrm{C}_{0}+^{8}\textrm{C}_{1}+^{8}\textrm{C}_{2}+^{8}\textrm{C}_{3}+............+^{8}\textrm{C}_{8}=2^{8}$ 

But according to question there is an edge with $8$ vertices , i.e. $^{8}\textrm{C}_{2}$

So, answer should be $\frac{^{8}\textrm{C}_{2}}{2^{8}}$


If they asking for Expected value, then there must be a weight or cost (like 1/2 given in GATE question). But, here that is not the scenario.

https://gateoverflow.in/1535/gate2013-24

https://www.quora.com/What-is-the-difference-between-probability-and-expectation

1 votes
1 votes
As said the answer is 1

But it's not right as the probability will be 2^28 -1 / 2^28 which is close to 1 but not one. The question does not specify Whether the graph chosen will be null or not thus total possibilities =2^28, only.one possibility when edge will not exist if it becomes null graph . So favourable events =2^28-1

. Thus probability becomes 2^28-1/28

Edit its 2^28 not 28
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