Question

enter the correct answer.
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)

P(A) = number of green marbles / total marbles = $\frac{5}{12}$

Step2: Calculate P(B after A)

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

Step3: Compute P(A then B)

Multiply P(A) and P(B after A): $\frac{5}{12} \times \frac{4}{11} = \frac{20}{132} = \frac{5}{33}$

Answer:

$\frac{5}{33}$