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token numbered

by sanju09 » Fri Feb 19, 2010 2:14 am
A bag contains token numbered 1 through 100. Two tokens were drawn randomly out of the bag. What is the nearest probability that the sum on it is more than 100?
(A) 0.505
(B) 0.515
(C) 0.525
(D) 0.535
(E) 0.545
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by ajith » Fri Feb 19, 2010 2:50 am
sanju09 wrote:A bag contains token numbered 1 through 100. Two tokens were drawn randomly out of the bag. What is the nearest probability that the sum on it is more than 100?
(A) 0.505
(B) 0.515
(C) 0.525
(D) 0.535
(E) 0.545
I can see that this is a sum of conditional probabilities

Total probability = probability that the sum is over 100 given that first num is one* probability that first num is one + ......

= 1/100*1/100+1/100*2/100+......... +1/100*100/100
= 1/100*(1+2+...100)/100
[spoiler]= 0.505

Yay! It gives me such a pleasure![/spoiler]
Last edited by ajith on Fri Feb 19, 2010 3:07 am, edited 2 times in total.
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by harsh.champ » Fri Feb 19, 2010 2:56 am
sanju09 wrote:A bag contains token numbered 1 through 100. Two tokens were drawn randomly out of the bag. What is the nearest probability that the sum on it is more than 100?
(A) 0.505
(B) 0.515
(C) 0.525
(D) 0.535
(E) 0.545
Case1:-1 is taken -Only 1 case (2nd token should be 100)
Case 2:-2 is taken - 2 cases(100 and 99)
|
|
|
This will continue upto
Case 50:-(51,52...100)50 no.s can be taken.

So,we have 1+2+3+---+49+50=1275

Total selections = 4950(100C2)

Hence ,the answer should be 0.2575
I dont know where I went wrong.
Doubling my answer will give 0.515 which is B.
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by sanju09 » Fri Feb 19, 2010 3:07 am
[spoiler]IMO A[/spoiler] B-)

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by vijay_venky » Sat Feb 20, 2010 1:47 am
Exact same procedure as harsh. But I think there needs to be some addition to it.

case - 51:52,53,54,55,.........100:::::49
case - 52:53,54,55,56,..........100::::48
.
.
.
.
.case 99:100.....................................1

So in effect the equation should look like 2(1+2+3+......+49)+50=49x50+50=50x50

And total = 100x99/2 So probability= (50x50)/(50x99)=0.505
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by shashank.ism » Sat Feb 20, 2010 7:21 am
ajith wrote:
sanju09 wrote:A bag contains token numbered 1 through 100. Two tokens were drawn randomly out of the bag. What is the nearest probability that the sum on it is more than 100?
(A) 0.505
(B) 0.515
(C) 0.525
(D) 0.535
(E) 0.545
I can see that this is a sum of conditional probabilities

Total probability = probability that the sum is over 100 given that first num is one* probability that first num is one + ......

= 1/100*1/100+1/100*2/100+......... +1/100*100/100
= 1/100*(1+2+...100)/100
[spoiler]= 0.505

Yay! It gives me such a pleasure![/spoiler]
Ajith I don't get your solution ....I think you have posted a good and shortest one but I m not getting.Please explain in bit detail..I really need to know about conditional probabilities...
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by harsh.champ » Sat Feb 20, 2010 8:06 am
vijay_venky wrote:Exact same procedure as harsh. But I think there needs to be some addition to it.

case - 51:52,53,54,55,.........100:::::49
case - 52:53,54,55,56,..........100::::48
.
.
.
.
.case 99:100.....................................1

So in effect the equation should look like 2(1+2+3+......+49)+50=49x50+50=50x50

And total = 100x99/2 So probability= (50x50)/(50x99)=0.505
Hey venky,
Thanks for correcting my mistake.
When I was solving this question ,I had thought that after 50 the same cases will repeat as the previous ones[Ex:-If 1st token is 45, the 2nd token can be 56,57,58,.......,100 so I had thought that when I consider 56 as the 1st taken and 45 as the 2nd token the same cases will repeat.]
Hence,I had stoppedthe soln. at 50.
But the numbers greater than 56 can be selected 57,58,59.....100 - I did not take this into consideration.
Good question for practice though.
It takes time and effort to explain, so if my comment helped you please press Thanks button :)



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by ajith » Sat Feb 20, 2010 9:12 am
shashank.ism wrote:
Ajith I don't get your solution ....I think you have posted a good and shortest one but I m not getting.Please explain in bit detail..I really need to know about conditional probabilities...
getting 1 in first , getting 2 in first ...., getting 100 in first are 100 mutually exclusive events each having a probability of 1/100. Now, given that I got 1 in first throw, the probability that I will cross 100 with the second throw is 1/100,given that I got 2 in first throw, the probability that I will cross 100 with the second throw is 2/100....

So total probability is the sum of all conditional probability

P(A) = P(B1/A1) * P(A1)+ P(B2/A2)*P(B2)....

= 1/100*1/100 + 2/100*1/100 +...... +100/100*1/100
= (1+2+3+..100)/(100*100)

= 5050/10000
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