There are 9 numbers (1-9) placed in a bag, if two numbers are drawn without replacement in succession, find the probability that the sum of the numbers is odd?
For the sum to be odd, one choice must be even , the other odd
(the sum of two odds is even, the sum of 2 evens is even)
We have 5 odds and 4 evens in the bag
so 2 cases:
prob(E, then O) = (4/9)(5/8) = 20/72 = 5/18
prob(O, then E) = (5/9)(4/8) = 5/18
prob(your event) = 5/18+5/18 = 10/18 = 5/9
check:
two other cases:
Prob(E,E) = (4/9)(3/8) = 1/6
prob(O,O) = 5/9)(4/8) = 5/18
sum: 5/9 + 5/18 + 1/6 = 1 , YEAHH