Probability confusion

#1
Hi guys, I have a question here. I understand the reasoning behind the solution, but I am also wondering about why an alternative solution is not correct.

Seven students are called on at random on two separate occasions. One of the students has been called on for the first occasion. What is probably that this student will NOT be called on the second time?

The correct way to solve this (1/7) for the first time and (1/7) for the second time. The chances of the student being called on twice is (1/49), so 1-(1/49) is 48/49.

The student has a 48/49 chance of NOT being called on twice.

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However, I was wondering why the following solution doesn't work--if you were to use (1/7) for the first time BUT (6/7) the second time to represent the chances of another student being called on. (1/7)*(6/7) = 6/49.

So the first time, the student has a 1/7 chance of being called on. The second time, he has a 6/7 chance of NOT being called on. Why can't we use this to establish the chances of him being called on only once?
 

BGM

TS Contributor
#2
1 - (1/7)^2 = probability of not being called twice
(1/7)(6/7) = probability of being called on the first time but not the second time
(6/7)(1/7) = probability of being called on the second time but not on the first time
(2C1)(1/7)(6/7) = probability of being called exactly once

Are these what you want?
 
#3
Thanks for the response. I am just confused as to why "probability of not being called twice" is not the same conceptually as "probability of being called on the first time but not the second time."

Doesn't each of these conditions mean the same thing? As in the "probability of being called on only once?" Why are their equations different?
 

BGM

TS Contributor
#4
Let
A be the event that he is not being called on both the first and the second time
B be the event that he is being called on the first time but not the second time
C be the event that he is being called on the second time but not on the first time
D be the event that he is being called on both the first and the second time

Then {A, B, C, D} forms a partition, i.e. exactly one of them will happen

Also P(A) = (6/7)(6/7), P(B) = (1/7)(6/7), P(C) = (6/7)(1/7), P(D) = (1/7)(1/7)
by independence

The event "not (being called twice)" = Ω\D = A∪B∪C
The event "being called on the first time but not the second time" = B
The event "being called only once" = B∪C

If you know the Binomial distribution well, let N be the number of times he being called
Then N ~ Binomial(2, 1/7)
and {N = 0} = A, {N = 1} = B∪C, {N = 2} = D