Probability

Concept:The probability of selecting a white ball changes after each draw because two additional balls of the same colour are added.Explanation:• Initially, white = 3, red = 5, total = 8.• $P(\text{white on 1st}) = \frac{3}{8}$.• After drawing a white, it is replaced and 2 more whites are added. So white becomes $3+2=5$, red remains 5, total = 10.• $P(\text{white on 2nd} \mid \text{white on 1st}) = \frac{5}{10} = \frac{1}{2}$.• After second white, white becomes $5+2=7$, red remains 5, total = 12.• $P(\text{white on 3rd} \mid \text{white on 1st and 2nd}) = \frac{7}{12}$.• Required probability = $\frac{3}{8} \times \frac{1}{2} \times \frac{7}{12} = \frac{21}{192} = \frac{7}{64}$.Answer:$\frac{7}{64}$ (Option A).