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
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
