Suppose you are playing a game of chance. If you bet $6 on a certain event, you will collect $174
(including your $6 bet) if you win. Find the odds used for determining the payoff.
A) 174 : 180
B) 1 : 28
C) 29 : 1
D) 28 : 1

A set consist of 2n-1 element. What is the number of subsets of this set which contain at most n-1 elements?

A. 2^(2n-2)....
B. 2^(2n) - 2
C. 2^(2n) -1
D. 2^(2n)
E. 2^(2n-1)

Could someone please help with the above questions. thanks!

I think for q2 it should b ....(2n-1)C0+(2n-1)C1+..........+(2n-1)C(n-1)
As we have to form subsets nd these subsets can have elements ranging from 0 to n-1
But I'm stuck with the last term......

hey your solution helps I think. In a 2n-1 binomial expansion there will be 2n terms ( ie 2n-1C0 to 2n-1C2n-1) but in our expansion we need the terms only uptil n-1, like you've done above- so that makes it half of the total number of terms in the binomial expansion. since the entire expansion= 2^2n-1, the first half of the expansion will be equal to (2^2n-1)/2 which gives 2^2n-2. Right? the answer is a) and i think this logic is correct. please verify.

no coz the zero term is also there na which makes it 2n terms actually and thats why half of those 2n terms in n terms-(0 to n-1 ). Also, any suggestions on the first question as well?

@Deepak
the answer according to my calculation is 120/343. For none of them to have birthdays on the same day- it should be 7.6.5.4/7.7.7.7= 120/343. please let me know if the answer is right.

@arushi ...I think I got the notion behind this ques.....let A be the event of winning then P(A)+P(A')=1
Then the in favour of winning is given by p(A)/p(A')
Now in terms of payoff ....the payoff when event A occurs is 174-6=168! Because we are considering payoff not total profit
The payoff associated with not winning to the gambler other side is 6. So,
Payoff(A)/payoff(A')
168/6=28:1
Which is read as if u bet with 6 nd if u win u'll get 168 as payoff and the betting amnt of 6 nd if u loose the gambler will take ur betting amnt ie 6

@ arushi
1st question
you bet 6 and after winning you get 174. By logic, in any game of chance payoff are kept in that way that u can't win in long run(otherwise casinos won't be profitable). Since, in this case if every chance u are betting 6
so if u bet 174/6 = 29 times ur net profit would be zero (u will win 1 time and loose 28 times) in long term that means ur probability to win is 1/29. So, for casino odds for determining payoff would be (odds favoured/ odds against)= 28/1.
and for me odds for winning would be 1/28 while probability for winning is 1/29.

2nd question
answer would be 2^(2n-2).
u have to divide the sum of series of (2n-1)C0+(2n-1)C1+..........+(2n-1)C(2n-1) by 2 .
sum of series = 2^(2n-1) and now divide by 2.
(2^(2n-1))/2= 2^(2n-2).

@Arushi
Sorry I don't have the answer to the question I have mentioned.
but,the ques mentions that no two have on same day,does this mean that 3 can have on same day and all can have on same day.