Initially half of the balls are odd. The first ball being odd has a probability of 1/2. Taking one odd ball out leaves 4 odd and 5 even left in the bag, what is the probability of taking an odd ball out now? How do you combine the probabilities of two different events?
Hello, mathewnaren!
Another approach . . .
Ten balls numbered 0, 1, 2, ... , 9 are in a box, and two balls are taken at random.
The probability that both numbers are odd is equal to:
. . $\displaystyle (A)\;\tfrac{1}{3}\qquad(B)\;\tfrac{1}{9} \qquad (C)\;\tfrac{4}{9}\qquad(D)\;\tfrac{2}{9} \qquad (E) \;1$
There are: $\displaystyle _{10}C_2 \,=\,45$ possible outcomes.
There are: $\displaystyle _5C_2 \,=\,10$ ways to get two odd numbers.
Therefore . . .