Question

a bag contains 12 marbles: 7 yellow, 5 green. if 1 marble is drawn, set aside, then another 1 is drawn, what is the probability of drawing a green marble both times?
event a: green marble on 1st draw
event b: green marble on 2nd draw
p(a, then b) = ?
give your answer as a fraction in simplest form.
compound probability
dependent events
p(a then b) =
p(a)·p(b after a)

Explanation:

Step1: Calculate P(A)

Total marbles = 12, green marbles = 5. So $P(A) = \frac{5}{12}$.

Step2: Calculate P(B after A)

After drawing 1 green marble, remaining marbles = 11, remaining green marbles = 4. So $P(B \text{ after } A) = \frac{4}{11}$.

Step3: Compute P(A then B)

Multiply the two probabilities: $P(A) \cdot P(B \text{ after } A) = \frac{5}{12} \cdot \frac{4}{11} = \frac{20}{132}$.

Step4: Simplify the fraction

Divide numerator and denominator by 4: $\frac{20 \div 4}{132 \div 4} = \frac{5}{33}$.

Answer:

$\frac{5}{33}$