Question

a box contains 4 red balls and 6 blue balls. two balls are drawn without replacement. what is the probability that both balls are red?

Explanation:

Step1: Total balls calculation

Total balls = 4 red + 6 blue = 10

Step2: Probability first red ball

$P(first\ red) = \frac{4}{10}$

Step3: Probability second red ball (no replacement)

Remaining red balls = 3, remaining total balls = 9, so $P(second\ red) = \frac{3}{9}$

Step4: Combined probability

Multiply the two probabilities: $\frac{4}{10} \times \frac{3}{9} = \frac{12}{90} = \frac{2}{15}$

Answer:

$\frac{2}{15}$