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Probability

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Probability

by radhika108 » Wed Apr 28, 2010 7:54 pm
There are 7 students in a class. One student is picked to answer a question. What is the probability that the same student is not picked the second time to answer another q?

The answer was worked worked out as 1- p (same student being picked twice)= 1-1/49= 48/49 I would like to understand the long method, so I can get concepts right. I am trying to work it out like:
It does not matter who you pick for 1st student, second student could be 1/6. But the answer is not right. Please explain where i am understandign it wrong? thanks in advance.

The other q was the sum of first 50 even numbers- this acc to me includes 0-98, and hence should be 50*49 ,but I see 50*51 as common nswer as 0 is not taken to account and they consider numbers 2-100
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by ajith » Wed Apr 28, 2010 10:40 pm
radhika108 wrote:There are 7 students in a class. One student is picked to answer a question. What is the probability that the same student is not picked the second time to answer another q?

The answer was worked worked out as 1- p (same student being picked twice)= 1-1/49= 48/49 I would like to understand the long method, so I can get concepts right. I am trying to work it out like:
It does not matter who you pick for 1st student, second student could be 1/6. But the answer is not right. Please explain where i am understandign it wrong? thanks in advance.
The answer to this question is not 48/49

There are 7*7 =49 ways to pick 2 students for questioning, in 7 of those same student gets picked twice.
So the probability you are looking for is (49-7)/49 = 6/7

In your way of reasoning, there are 7 ways to pick the second student out of which 6 are favorable the probability is 6/7
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by ajith » Wed Apr 28, 2010 10:42 pm
radhika108 wrote:The other q was the sum of first 50 even numbers- this acc to me includes 0-98, and hence should be 50*49 ,but I see 50*51 as common nswer as 0 is not taken to account and they consider numbers 2-100
Technically, negative even numbers are also even numbers, so they have no start and no end. You need one more qualifier - First 50 positive/natural/whole even numbers - Depending on that the answer varies. Be assured that GMAT will specify what exactly they want.
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by brijesh » Thu Apr 29, 2010 2:53 am
I think the answer is 42/49.

Method-1

Total no of ways of sellecting 2 studensts= 7*7
Favourable no of ways = 7*6 (second time the second student was not sellected).
Probabality= 42/49

Method-2

Both the time same student is sellected= 7*1
Probabality= 1-7/49= 42/49
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by radhika108 » Thu Apr 29, 2010 5:28 am
Thanks ajith and brijesh for your answers.

42/49 is 6/7.
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