No Talking!

Let A, B, C, D, E denote each person respectively. So there's 5 ways to select 2 of them in the group. Which is a total of nCr(5, 2) = 10 ways.

They are: AB, AC, AD, AE, BC, BD, BE, CD, CE, DE

Suppose person A will never be asked to speak in the first week, then his/her probability of not asked is 6/10

And the probability of not getting chosen for the following week is (6-1)/(10-1)

And the following week: (6-2)/(10-2)

And the last week (6-3)/(10-3)

Multiply these probabilities together gives 1/14

Since there's a total of 5 people, the probability ANY person not asked is 5 times the probability calculated above.

Hence, the desired probability is 5/14

Because gcd(5,14) = 1, then a = 5, b = 14. So a + b = 19

selcting a person 5c1 ways then the probability that the person doesnt speak is (4c2x4c2-1x4c2-2x4c2-3)/(5c2x5c2-1x5c2-2x5c2-3) = 1/14 so the net probability becomes 5c1x(1/14) =5/14 so a+b=5+14=19

To begin, choose one person to be excluded from speaking for all four weeks. Initially, there are $6$ of $10$ pairings that do not include this person, and each week, as one of these pairings is selected, both the number of exclusive pairings remaining and the number of total pairings remaining decrease by one.

Thus, the probability that all occurring pairings exclude the chosen person is $\frac{6}{10} * \frac{5}{9} * \frac{4}{8} * \frac{3}{7} = \frac{1}{14}$ .

However, that only considers one person for exclusion, while there are five, so the total probability is $5 * \frac{1}{14} = \frac{5}{14} = \frac{a}{b}$ $a + b = 5 + 14 = \boxed{19}$

If there were 2 people who never spoke, then only 3 of the people would speak, but there are only $\binom{3}{2} = 3$ pairs of people, so the same pair would have to speak on two different weeks, which is not allowed. Thus, at most one person never speaks.

Consider one person, $P$ in this group. Since there are 5 people in the group, there are $\binom{5}{2} = 10$ pairs of people. 4 of these pairs contain $P$ . The probability that $P$ will not be selected in the first week is $\frac{6}{10}$ . In the second week there is one less pair of people that can be selected, so the probability that $P$ will not have to speak is $\frac{5}{9}$ . For the third and fourth weeks, the probability will be $\frac{4}{8}$ and $\frac{3}{7}$ respectively. So the probability that $P$ will not speak any week is $\frac{6 \times 5 \times 4 \times 3}{10 \times 9 \times 8 \times 7} = \frac{1}{14}$ .

Since there are 5 people in the group, we can calculate the probability that one of them never speaks using the principle of inclusion and exclusion. We showed above that there is never a case where 2 of them do not speak, so the probability is just $5 \times \frac{1}{14} = \frac{5}{14}$ . So $a + b = 5 + 14 = 19$ .

Firstly, let's determine in how many ways 2 people out of 5 can be selected for four times.

Denoting each of them as A,B,C,D,E ; total (5C2)C4 ways are there to select.

Now, excluding one person, out of 4 people we can select as the same way (4C2)C4. Excluding each of 5 people, total number of ways becomes 5*((4C2)C2).

Then, the probability is 5*((4C2)C2) / (5C2)C4 = 75/210 = 5/14

Let us consider 5 persons,i.e,A,B,C,D,E.Now,let us choose that particular person(who will never be asked to speak ),so we get 5 possibilities.Now considering other 4,for 1st week we can select by 4C2=6 methods.For 2nd one,we have to remove that previous pair,thus 5.Similarly,for 3rd-4 methods,and for last week-3.Now for ideal case,possibilities we can get for 1st week is 5C2,i.e=10.Following d rule (that no pair of people will speak together more than once),for 2nd week we get 9,for3rd-8,& for 4th-7.Thus final probability comes out to be =(total required possibilities)/(total ideal possibilities)=5(6 5 4 3)/10 9 8 7=5/14.

selecting a person 5c 1 ways then the probability that the person does'nt speak is (4c2 x 4c2 - 1x4c2 -2x4c2 - 3)/(5c2 x 5c2 - 1x5c2 - 2x5c2 -3 ) = 1/14 so the net probability becomes 5c1x(1/14) =5/14 so a+b=5+14=19