Probability distribution without replacement

[Guest]

I am stuck on the following Question. I worked out the answer for replacement but I am completely lost on the "without replacement" aspect

A bag contains five tokens numbered 2, 3, 6, 7, and 8.
Two tokens are taken in succession out of the bag without replacement.
a) Create the probability distribution for “x” being the number of odd numbered
tokens drawn.

[HallsofIvy]

The numbers are 2, 3, 6, 7, 8. There are 2 odd numbers, 3 and 7, and 3 are even, 2, 6 and 8.

The probability the first number is even is 3/5 and then, without replacement, there are 4 numbers left, 2 of which are even. The probability that the second number is also even is 2/4= 1/2. The probability that both numbers are even, so that there are 0 odd numbers, is (3/5)(1/2)= 3/10.

The probability the first number is odd is 2/5 and then there are 4 numbers left, 3 of which are even so the probability the second number is even is 3/4. The probability the numbers are odd and even, in that order is (2/5)(3/4)= 3/10 and the number of odds is 1.

The probability the first number is even is 3/5 and then there are 4 numbers left of which 2 are odd so the probability the second number is odd is 2/4= 1/2. The probability the numbers are even and odd, in that order, is (3/5)(1/2)= 3/10 again.

The probability that the number are "even and odd" or "odd and even", so that the number of odds is 1, is 3/10+ 3/10= 6/10= 3/5.

The probability the first number is odd is 2/5 and then there are 4 numbers left, of which 1 is odd so the probability the second number is also odd is 1/4, so that the probability the number of odds is 2, is (2/5)(1/4)= 1/10.

P(0)= 3/10, P(1)= 6/10= 3/5, P(2)= 1/10.

As a check, note that the sum of these is 3/10+ 6/10+ 1/10= 10/10= 1.