2. ## Re: 10 Balls probability

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?

3. ## Re: 10 Balls probability

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:

. . $(A)\;\tfrac{1}{3}\qquad(B)\;\tfrac{1}{9} \qquad (C)\;\tfrac{4}{9}\qquad(D)\;\tfrac{2}{9} \qquad (E) \;1$

There are: $_{10}C_2 \,=\,45$ possible outcomes.

There are: $_5C_2 \,=\,10$ ways to get two odd numbers.

Therefore . . .