How do you calculate the probability of picking two blue balls from a bag without replacement?

To calculate the probability of picking two blue balls from a bag without replacement, use the formula for dependent events.

First, let's break down the problem. Suppose you have a bag containing a mix of blue and other coloured balls. The total number of balls in the bag is \( n \), and the number of blue balls is \( b \).

The probability of picking a blue ball on the first draw is simply the number of blue balls divided by the total number of balls, which is \( \frac{b}{n} \).

Once you have picked the first blue ball, you do not replace it. This means the total number of balls in the bag is now \( n-1 \), and the number of blue balls left is \( b-1 \).

The probability of picking a blue ball on the second draw, given that the first ball was blue, is \( \frac{b-1}{n-1} \).

To find the overall probability of both events happening (picking two blue balls in a row), you multiply the probabilities of each individual event. Therefore, the probability \( P \) is:

\[ P = \frac{b}{n} \times \frac{b-1}{n-1} \]

For example, if you have a bag with 5 blue balls and 10 balls in total, the probability of picking two blue balls without replacement would be:

\[ P = \frac{5}{10} \times \frac{4}{9} = \frac{1}{2} \times \frac{4}{9} = \frac{4}{18} = \frac{2}{9} \]

So, the probability of picking two blue balls from this bag without replacement is \( \frac{2}{9} \).