Question

A group of 10 students is interested in a football match (with Adam, George, Cyril and David) and they received 3 tickets. They're going to draw for the tickets. What is the probability that the lucky ones are:
Adam and George or George and Cyril?

Answer 1

Note: the result is supposed to be \[\frac{ 1 }{ 8 }\] according to the book

Answer 2

Probabilities (if I recall correctly) are (desired outcomes) / (universe of total outcome possibilites)

You would have to build a fraction from there.

Answer 3

You are right. I wouldn't have a problem if it was for example Adam + George and Cyril + David, since then it would have become simply:
P(A) + P(B)
3/10 * 2/9 + 3/10 * 2/9 (the probability of Adam getting a ticket would be 3/10, George getting one 2/9, repeat)

However, I am completely lost how to do it in this case. I believe I am supposed to do:
P(A or B) = P(A) + P(B) - P(A and B)
However, I am not sure, whether is the P(A) and P(B) supposed to be done in the same way (3/10 * 2/9) and how to create the P(A and B)

Answer 4

"and" signifies multiplication of probabilities and "or" signifies addition of probabilities.

Answer 5

I am confused by the wording of the question. I think it is P(Adam)*P(George) + P(George)*P(Cyril) ???

Answer 6

But maybe it could P(George)*2P(Adam/Cyril)

Answer 7

George must be chosen then it could be adam or cyril next so you there is double the chance times george's probability

Answer 8

Ok, I believe I have found the solution, in case anyone was wondering.
This is how it's built
P(A) = 3/10 * 2/9
P(B) = 3/10 * 2/9
P(A and B) = 3/10*2/9*1/8 - since that's chance of all of this happening at the same time

P(A or B) = 3/10*2/9 + 3/10*2/9 - 1/120
P(A or B) = 1/8

Thanks for your help!

Answer 9

Good job, I probably would have got it wrong.