Question

a bag contains 8 marbles: 5 red, 3 blue. if 1 marble is drawn, set aside, then another 1 is drawn, what is the probability of drawing a red marble both times? event a: red marble on 1st draw event b: red 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)

$P(A) = \frac{\text{Number of red marbles}}{\text{Total marbles}} = \frac{5}{8}$

Step2: Calculate P(B after A)

After drawing 1 red marble, remaining red marbles = 4, total marbles = 7. So $P(B \text{ after } A) = \frac{4}{7}$

Step3: Compute P(A then B)

$P(A \text{ then } B) = P(A) \times P(B \text{ after } A) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56}$

Step4: Simplify the fraction

$\frac{20}{56} = \frac{5}{14}$ (divide numerator and denominator by 4)

Answer:

$\frac{5}{14}$