A hat contains 5 tickets numbered 1,2,3,4,5. Two tickets are drawn at random from the hat. Find the chance that the number on the two tickets differ by two or more if the draws are made
You didn't finish your question "if the draws are made..."
Made what?
Are the draws with replacement or without?
With replacement, sometimes it is easier with these kinds of problems just to enumerate the choices.
If you draw a 1 (1/5) probability, you have a 3/5 (3,4,5) chance that the numbers will be two or more apart. So you multiply these two events to get the probability. The rest of the probabilities are similar.
1: (1/5) * (3/5)
2: (1/5) * (2/5)
3: (1/5) * (2/5)
4: (1/5) * (2/5)
5: (1/5) * (3/5)
Since each of these events are independent, you add them all together, giving 12/25 or a bit less than 1/2.
Without replacement, the same idea applies but now you remove one the card you drew, making it easier to get one two or more apart:
1: (1/5) * (3/4)
2: (1/5) * (2/4)
3: (1/5) * (2/4)
4: (1/5) * (2/4)
5: (1/5) * (3/4)
Giving 12/20 = 3/5, or a bit more than 1/2.
Hello, billy!
A hat contains 5 tickets numbered 1,2,3,4,5.
Two tickets are drawn at random from the hat.
Find the probability that the numbers on the two tickets differ by two or more.
With such a small number of outcomes, we can list all of them.
If the tickets are drawn with replacement, the outcomes are:
. . $\displaystyle \begin{array}{ccccc}(1,1) & (1,2) & (1,3) & (1,4) & (1,5) \\ (2,1) & (2,2) & (2,3) & (2,4) & (2,5) \\
(3,1) & (3,2) & (3,3) & (3,4) & (3,5) \\ (4,1) & (4,2) & (4,3) & (4,4) & (4,5) \\ (5,1) & (5,2) & (5,3) & (5,4) & (5,5) \end{array}$ . . 25 outcomes
If they are drawn without replacement, the outcomes are:
. . $\displaystyle \begin{array}{ccccc} & (1,2) & (1,3) & (1,4) & (1,5) \\ (2,1) & & (2,3) & (2,4) & (2,5) \\
(3,1) & (3,2) & & (3,4) & (3,5) \\ (4,1) & (4,2) & (4,3) & & (4,5) \\ (5,1) & (5,2) & (5,3) & (5,4) & \end{array}$ . . 20 outcomes