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In a lottery, 10 tickets are drawn at random out of 50 tickets numbered from 1 to 50. What is the expected value of the sum of numbers on the drawn tickets?
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255?
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yes! @aditya333 can you please explain the solution in step by step in a detailed manner

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I am also getting 255. But when dividing by 401 not by probability.

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see my comment on the answer..Is it correct?

we have no's from 1 to 50. So the minimum sum that can be obtained by picking 10 numbers is 1+2+3...10= 55

maximum sum that can be obtained is 41+42+43+.....50=455

so our range of sum is from 55 to 455

now lets assume that no of ways to pick 10 numbers from these 50 numbers is S

We have to calculate the probabilities of the respective sums

so for 55 there is only 1 way of picking 10 numbers such that their sum is 55

similarly there is only 1 way of picking 10 numbers such that the sum is 455

now we have to calculate $\sum x*f(x)$ where x is the sum and f(x) is the respective probability

now the denominator of this summation will S(as assumed earlier). we only need to worry about the numerator

in the numerator the terms would be like 1*55 + 1*56 +.............+1*454 +1*455. now this sequence is symmetric .So lets pair up the numbers like(1st and last, 2nd and 2nd last ...)  so we will get something like 1*510 +1*510+........ (NOTE : we will have terms like 2*510 ,3*510 etc also)

now if we observe closely this would sum up to 510*(S/2).

so our final answer would be (510*(S/2))/S= 510/2 =255

If this is still not clear how S/2 comes take up smaller examples and see for urself

by Junior (547 points)
edited
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well explained
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from 55,56,..,455, there are 401 terms.

So the 201th term cant be paired up with anyone.

So pairs are (55,455), (56,454),...(254,256),255.

There are a total of 200 pairs.

Adding each pair up, we get sum as 510.

So expectation will be [(510*200)+255]/401 = 255.
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How will u find probability of sum 55?
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@Shubhanshu Probability of each of the sums will be 1/401 right? Because there is only one way to get each sum by adding up 10numbers.

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But to get the sum equal to 55. You have to select $1,2,.....,10$ tickets exactly. So its Probability is $\frac{favourable}{Total}$ and here favorable occurs when exactly $1,2,.....,10$ tickets gets selected which means only 1 way and Total is $C(50,10)$ which means all possible combination with 10 tickets. So, in total it should be

$x*P(X = x) \rightarrow 55*\frac{1}{C(50,10)}$
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@Somoshree Datta 5 no probability of every sum is not same. First the probability will increase and then decrease .The sequence is symmetric

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@Shubhanshu

Yes.. But then how are you arriving at the answer? I'm not getting the explanation.. Can u pls explain?

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oops i have written 550 .thanks for pointing out
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@Somoshree Datta 5 here the sum of the coefficients of each um terms equals to S,ie(the total no of ways of picking 10 numbers from 50 numbers) . we have a total of 455-55+1=401 terms and the middle sum term is  255.

as the number of terms is odd and the middle term  wont pair with any term so let its coefficient be a so the numerator sum becomes (S-a)/2 + a*255  (we can write this because the sequence is symmetric). so a*255 can be written as (510*a)/2  s0 the numerator becomes (S-a)/2 + 510*a/2= (S*510)/2

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@aditya333 for selecting sum = 55, you have to select 1 to 10 tickets only, so here favourable cases should be $C(10,10)$ and total cases would $C(50,10)$. So, the probability of each individual sum = $\frac{C(10,10)}{C(50,10)}$ And as per expectation it would be $E[X] = \sum X*P(X)$ so here $X = 55$ and $P(X) = \frac{C(10,10)}{C(50,10)}$.

Where I am doing wrong?

+1

@Shubhanshu consider this example.Suppose we have to select 2 numbers from  1 to 6

so sum would range from 3 to 11

the sum table is shown below

SUM        FAVOURABLE_CASES

3               (1,2)

4                  (1,3)

5                  (1,4), (2,3)

6                    (1,5) ,(2,4)

7                     (1,6) ,(2,5),(3,4)

8                    (2,6) (3,5)

9                     (3,6) ,(4,5)

10                     (4,6)

11                      (5,6)

so summing the favourable cases we get 15 which is nothing but 6C2

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And total cases for your example?
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total cases =15
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so what is S here?

Is it C(50,10)??

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yeah
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For every outcome between 55 to 455, is the probability of getting any outcome 1/50?
expected value of the sum on one number= ((50x51)/2) x (1/50) = 25.5

for 10 numbers =25.5x10=255
by (85 points)