# probability doubts

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## probability doubts

 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!
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## Re: probability doubts

 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......
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## Re: probability doubts

 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.
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## Re: probability doubts

 But half of 2n-1 isn't n-1.....den y hv u taken it as half...?
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## Re: probability doubts

 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?
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## Re: probability doubts

 Oh yeah...total no .of terms will b 2n-1+1.... Thanks.... I m also stuck at q1
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## Re: probability doubts

 a group of 4 friends celebrate their birthday in a particular week.what is the probability that no two friends celebrate their birthdays on same day?? 1.39/343 2.60/2401 3.60/343 4.120/343
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## Re: probability doubts

 In reply to this post by Arushi @Arushi quest 1... Since we are playing a game of chance there is only 1 way we can win.. We divide 174/6=29 hence ans 29:1
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## Re: probability doubts

 hey-the answer is 28:1 but I still cant understand the logic of the question.
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## Re: probability doubts

 In reply to this post by Deepak_Ka @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.
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## Re: probability doubts

 In reply to this post by Arushi my mistake total is 29 and favourable is1...so28:1..
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## Re: probability doubts

 In reply to this post by Arushi @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
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## Re: probability doubts

 In reply to this post by Arushi @ 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). i think it helps.