A jar contains 7 balls numbered from 1 to 7

• November 29th 2010, 02:40 AM
Narek
A jar contains 7 balls numbered from 1 to 7
A jar contains 7 balls numbered from 1 to 7. If a ball is selected at random, and the number is noted down, and a second ball is selected without replacing the first, what is the probability that the two-digit number is even?

My Solution: if the number has to be even then its last digit is either 2,4,6.
Odd Set = {1,3,5,7}
Even Set = {2,4,6}
There are two possibilities to have our desired outcome:

I) Either both balls selected are even numbers therefore (C(1,3)*C(1,2)) / 7*6 = 6/42

II) The first one is odd but the other is even therefore (C(1,4)(1,3)) / 7*6 = 12/42

The result is : 6/42 + 12/42 = 18/42

Is this correct? Is there any easy way? Thank you
• November 29th 2010, 02:47 AM
dwsmith
I would think you would just do:

$\displaystyle \frac{4}{7}*\frac{1}{2}=\frac{4}{14}+\frac{3}{7}*\ frac{1}{3}=\frac{2}{7}+\frac{1}{7}=\frac{3}{7}$

The end results are the same but the odd results differ.
• November 29th 2010, 03:12 AM
Narek
I'm sorry but I don't understand. Would you please explain so I can understand the concept?
• November 29th 2010, 03:16 AM
dwsmith
Never mind. I mixed up what you said was odd and even sorry.